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2021中考-选择填空满分限时训练3

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2021年中考数学一轮复习选择填空提分特训(3) 附答案

2021年中考数学一轮复习选择填空提分特训(3) 附答案

选择填空提分特训(三)[限时:30分钟满分:48分]一、选择题(每小题3分,共36分)1.下列四个数中,最小的正数是()A.-1B.0C.1D.22.某种植物细胞的直径约为0.00015 mm,用科学记数法表示数0.00015为()A.1.5×104B.15×10-3C.1.5×10-3D.1.5×10-43.下列运算正确的是()A.x3·x2=x6B.x2+x2=x4C.(3x2)2=6x4D.x(x-1)=x2-x4.如图X3-1是由几个大小相同的小正方体搭成的几何体的俯视图,小正方形中的数字表示在该位置上小正方体的个数,则该几何体的左视图是()图X3-1图X3-25.将抛物线y=3x2向上平移3个单位长度,再向左平移2个单位长度,那么得到的抛物线的解析式为()A.y=3(x+2)2+3B.y=3(x-2)2+3C.y=3(x+2)2-3D.y=3(x-2)2-36.如图X3-3,AB∥CD,下列关系式中成立的是()图X3-3A.∠1=∠3B.∠2+∠3=180°C.∠2+∠4<180°D.∠3+∠5=180°7.在一个不透明的口袋中,装有一些除颜色外完全相同的红、白、黑三种颜色的小球,已知袋中有红球5个,白球,则袋中黑球的个数为()23个,且从袋中随机摸出一个红球的概率是110A.27B.23C.22D.188.下列命题正确的是()A.对角线互相垂直且相等的四边形是菱形B.两边及一角对应相等的两个三角形全等C.16的平方根是4D.一组数据2,0,1,6,6的中位数和众数分别是2和69.若式子√k-1+(k-1)0有意义,则一次函数y=(k-1)x+1-k的图象可能是()图X3-410.长沙市某机械厂四月份生产零件50万个,第二季度生产零件182万个.设该厂五、六月份平均每月生产量的增长率为x,那么x满足的方程是()A.50(1+x)2=182B.50(1+x)2+50(1+x)=182C.50+50(1+x)+50(1+x)2=182D.50(1+2x)=18211.如图X3-5,直径为10的☉A经过点C(0,5)和点O(0,0),B是y轴右侧☉A优弧上一点,则∠OBC的余弦值为()图X3-5A.12B.34C.√32D.4512.在▱ABCD中,AD=8,AE平分∠BAD交BC于点E,DF平分∠ADC交BC于点F,且EF=2,则AB的长为()A.3B.5C.2或3D.3或5二、填空题(每小题3分,共12分)13.在函数y=√x+2x中,自变量x的取值范围是.14.在一个不透明的口袋中,装有A,B,C,D4个完全相同的小球,随机摸取一个小球然后放回,再随机摸取一个小球,两次摸到同一个小球的概率是.15.“书法艺术课”开课后,某同学买了一包宣纸练习软笔书法,且每逢星期几写几张,即每星期一写1张,每星期二写2张,…,每星期日写7张.若该同学从某年的5月1日开始练习,到5月30日练习完后累积写完的宣纸总数超过120张,则可算得5月1日到5月28日他共用宣纸张数为,并可推断出5月30日应该是星期.16.如图X3-6,抛物线y1=12(x+1)2+1与y2=a(x-4)2-3交于点A(1,3),过点A作x轴的平行线,分别交两条抛物线于B,C两点,且D,E分别为抛物线y1与y2的顶点.则下列结论:①a=23;②AC=AE;③△ABD是等腰直角三角形;④当x>1时,y1>y2.其中正确的结论是(填序号).图X3-6附加训练17.解不等式组:{3(x+1)>2x+1,x+72>4x,并写出它的所有整数解.18.如图X3-7,点E是▱ABCD的CD边的中点,AE,BC的延长线交于点F,CF=3,CE=2,求▱ABCD的周长.图X3-7【参考答案】1.C2.D3.D4.D5.A6.D7.C [解析]设袋中黑球的个数为x ,则摸出红球的概率为523+5+x =110,所以x=22,故选C .8.D 9.B 10.C11.C [解析]设☉A 与x 轴的另一个交点为D ,连接CD , ∵∠COD=90°,∴CD 是直径,即CD=10. ∵C (0,5),∴OC=5, ∴OD=√CD 2-OC 2=5√3. ∵∠OBC=∠ODC ,∴cos ∠OBC=cos ∠ODC=OD CD =5√310=√32. 故选C .12.D [解析]如图①,在▱ABCD 中,∵BC=AD=8,BC ∥AD ,CD=AB ,CD ∥AB , ∴∠DAE=∠AEB ,∠ADF=∠DFC.∵AE 平分∠BAD 交BC 于点E ,DF 平分∠ADC 交BC 于点F , ∴∠BAE=∠DAE ,∠ADF=∠CDF , ∴∠BAE=∠AEB ,∠CFD=∠CDF , ∴AB=BE ,CF=CD.∵EF=2,∴BC=BE+CF -EF=2AB -EF=8, ∴AB=5;如图②,在▱ABCD中,∵BC=AD=8,BC∥AD,CD=AB,CD∥AB,∴∠DAE=∠AEB,∠ADF=∠DFC.∵AE平分∠BAD交BC于点E,DF平分∠ADC交BC于点F,∴∠BAE=∠DAE,∠ADF=∠CDF,∴∠BAE=∠AEB,∠CFD=∠CDF,∴AB=BE,CF=CD.∵EF=2,∴BC=BE+CF+EF=2AB+EF=8,∴AB=3.综上所述:AB的长为3或5.故选D.13.x≥-2且x≠014.1415.112五或六或日[解析]5月1日到5月28日共28天,包含4个完整的星期,∴他写的张数为:4×7×(1+7)2=112.若5月30日为星期一,所写张数为112+7+1=120,若5月30日为星期二,所写张数为112+1+2<120,若5月30日为星期三,所写张数为112+2+3<120,若5月30日为星期四,所写张数为112+3+4<120,若5月30日为星期五,所写张数为112+4+5>120,若5月30日为星期六,所写张数为112+5+6>120,若5月30日为星期日,所写张数为112+6+7>120,故5月30日可能为星期五、六、日.16.①③[解析]抛物线y2=a(x-4)2-3过点A(1,3),∴3=9a-3,解得a=23,故①正确;由题意可知E(4,-3),点A(1,3)与点C关于直线x=4对称,得到点C(7,3),∴AC=6,而AE=√(1-4)2+(3+3)2=3√5,故AC≠AE,②不正确;由抛物线的对称性可知AD=BD,易知点B的坐标为(-3,3).由点B(-3,3)和点D(-1,1)易知直线BD的解析式为y3=-x.由点A(1,3)和点D(-1,1)易知直线AD的解析式为y4=x+2.∴k1·k2=-1,∴AD⊥BD,故③正确;由12(x+1)2+1=23(x-4)2-3,解得x1=1,x2=37,所以当1<x<37时,y1>y2,故④错误.17.解:{3(x+1)>2x+1,①x+72>4x.②解不等式①,得x>-2,解不等式②,得x<1,∴不等式组的解集为-2<x<1,∴不等式组的所有整数解为-1,0. 18.解:∵四边形ABCD是平行四边形,∴AD∥BC,∴∠DAE=∠F,∠D=∠ECF.又ED=EC,∴△ADE≌△FCE(AAS),∴AD=FC=3,DE=CE=2,∴DC=4.∴▱ABCD的周长为2(AD+DC)=14.。

2021年内蒙古中考英语重点题型限时训练： 完形填空+ 阅读理解(A)+ 选词填空+ 任务型阅读

2021年内蒙古中考英语重点题型限时训练： 完形填空+ 阅读理解(A)+ 选词填空+ 任务型阅读

完形填空+ 阅读理解(A)+ 选词填空+ 任务型阅读Ⅰ.完形填空He has millions of students from all over the world. They listen to his classes online. They 1see his face on the screen. Salman Khan, 42, from the US, is the founder(创始人) of the learning website Khan Academy.Unlike other online courses that film teachers 2classes, Khan’s videos only use his voice and a digital blackboard for writing. He says he doesn’t want a human face to distract(使分心) students. Khan also makes his classes 310 to 15 minutes so that students won’t lose interest. They can also get scores from taking classes.The courses are translated into more than 36 different languages 4Chinese. About 80 teachers from the world’s top universities are now also teaching on the website. But all this started nine years ago with just one man—Khan. These kinds of special courses make Khan Academy one of the world’s 5online education websites.In 2006, Khan made homemade videos to help his 6grade cousin with math. He posted the videos online and got thousands of views. This encouraged him to make more videos and create his own website Khan Academy.Khan understands that many kids, especially those in developing countries, cannot get a good education. He has three degrees from universities. He wants to use his 7to help more people. Thus, his courses are all offered for free. “Our task is to 8 a free, world-class education for anyone, anywhere,” he said.Khan’s website has 9become very popular worldwide but won a lot of respect(尊重). Even Bill Gates, the founder of Microsoft, is one of his fans. Gates 10over $1.5 million to support Khan and named him “the teacher to the world”.()1.A.always uallyC.neverD.sometimes()2.A.to give B.givingC.giveD.gave()3.A.as long as B.as much asC.as short asD.as soon as()4.A.including B.withC.asD.beside()5.A.least popular B.more popularC.popularD.most popular()6.A.nine B.ninethC.ninthD.the ninth()7.A.knowledge B.fameC.courageD.value()8.A.give B.offerC.provideD.pay()9.A.ever B.alreadyC.notD.not only()10.A.gave out B.gave offC.gave upD.gave awayⅠ. 阅读理解The Digital Media Summer CampThe digital media(数字媒体) summer camp is a great chance for students aged 7 to 18 to have STEM courses. Students can learn to build an app, program a robot, or explore (探索) the world of music production. STEM courses will also help students create, learn teamwork and solve problems.Students will learn from our teachers who have been employed(聘用) for both their teaching experience and rich knowledge in their fields. Besides making new friends, students will leave with amazing final projects and lasting memories.Time: August 1st—August 31st Daily Schedule:8:30 a.m.—10:00 a.m. 10:30 a.m.—12:00 p.m. 12:00 p.m.—1:30 p.m. 1:30 p.m.—3:00 p.m. 3:30 p.m.—4:30 p.m.Hands-on STEM Instruction Continued Instruction & CreationLunch BreakRobots ProgrammingMusic Appreciation(鉴赏)Saturday Open House:At 3:30 p.m. every Saturday, parents are invited to our Open House. Students can show the projects they have been working on. Their parents can meet the teachers and explore more chances for their children to continue learning and creating.1.The digital media summer camp can help students .()A.keep healthyB.learn a foreign languageC.improve their writing abilityD.develop STEM knowledge2.Which word can best describe the teachers of the summer camp?()A.Experienced.B.Humorous.C.Young.D.Strict.3.Which of the following statements is NOT true?()A.Parents can be invited to see their kids’ projects.B.Students will leave the camp with amazing final projects.C.Students’ lunch break lasts two hours.D.Students can learn to program a robot in the summer camp.Ⅰ. 选词填空阅读下列短文,用方框内所给单词的正确形式填空。

2021年河南省中考数学总复习之限时专练选择填空题型部分

2021年河南省中考数学总复习之限时专练选择填空题型部分

2021年河南省中考数学总复习之限时专练选填题组限时专练(一)(时间:22~25分钟 分值:45分)一、选择题(每小题3分,共30分)1.若m 是-2 021的绝对值,则m 的值为( A ) A.2 021B.-2 021C.12 021 D.-12 021解析 ∵m 是-2 021的绝对值,∴m=2 021.故选A.2.2019年,我国国内生产总值约为99.09万亿元,稳居世界第二位,其中99.09万亿用科学记数法表示为( C ) A.0.990 9×1014B.99.09×1012C.9.909×1013D.9.909×1014解析 99.09万亿=9.909×1013,故选C.3.如图是正方体的一种展开图,其每个面上都有一个汉字,那么在原正方体中与“你”字相对面上的字是( C )A.中B.考C.顺D.利4.下列计算正确的是( D )A.3x-2x=1B.x(-x2)=x3C.x2÷x=2D.(-x3)2=x6解析 A.3x-2x=x,故此选项错误;B.x(-x2)=-x3,故此选项错误;C.x2÷x=x,故此选项错误;D.(-x3)2=x6,故此选项正确.故选D.5.(2020许昌二模)某班30位同学的安全知识测试成绩统计如下表所示,其中有两个数据被遮盖,下列关于成绩的统计量中,与被遮盖的数据无关的是( C )成绩(分) 24 25 26 27 28 29 30 人数■■ 3 3 6 7 9A.平均数,方差B.中位数,方差C.中位数,众数D.平均数,众数解析 由题表数据可知,成绩为24分、25分的人数和为30-3-3-6-7-9=2, 成绩为30分的人数最多,因此成绩的众数是30,将成绩按从小到大的顺序排列,处在第15、16位的两个数都是29,因此中位数是29, 因此中位数和众数与被遮盖的数据无关, 故选C.6.明代数学家程大位的《算法统宗》中有这样一个问题,其大意为:有一群人分银子,如果每人分七两,那么剩余四两;如果每人分九两,那么还差八两.设银子有x 两,共有y 人,则可列方程组为( B ) A.{7y =x +49y +8=x B.{7y =x -49y =x +8C.{7x +4=y 9x -8=y D .{7y =x +49y =x +87.关于x 的一元二次方程(x-1)·(x-3)=-x-2的根的情况为( D ) A.有两个不相等的实数根 B.有两个相等的实数根 C.无法确定 D.没有实数根解析方程可化为x2-3x+5=0,∴Δ=(-3)2-4×1×5=-11<0,∴原方程没有实数根,故选D.8.(2020郑州二模)在一个不透明的袋子里装有若干个白球和15个红球,这些球除颜色不同外其余均相同,每次从袋子中摸出一个球记录下颜色后放回,经过多次重复试验,发现摸到红球的频率稳定在0.6,则袋中约有白球( B )A.5个B.10个C.15个D.25个解析设袋中约有白球x个,根据题意得,15=0.6,解得x=10,15+x经检验,x=10是分式方程的解.则袋中约有白球10个.故选B.9.如图,在平面直角坐标系中,点A的坐标为(2,0),△OAB是等边三角形,一动点P从O点出发,以每秒1个单位长度的速度沿O→A→B→O→A→…做循环运动,那么第2 020秒时,点P的坐标为( A )A.(1,√3)B.(2,0)C.(12,√32) D.(-12,√32)解析由题意得,第1秒时,点P的坐标为(1,0); 第2秒时,点P的坐标为(2,0);第3秒时,点P的坐标为(2-1×cos 60°,1×sin 60°),即(32,√3 2);第4秒时,点P的坐标为(1,2×sin 60°),即(1,√3);第5秒时,点P的坐标为(12,√3 2);第6秒时,点P的坐标为(0,0);第7秒时,点P的坐标为(1,0),与第1秒时点P的坐标相同; ……故点P的坐标每6秒一个循环,∵2 020÷6=336……4,∴第2 020秒时,点P的坐标与第4秒时点P的坐标相同,为(1,√3),故选A.10.已知抛物线y=ax2+bx+c(a≠0)如图所示,它与x轴的两个交点分别为(-1,0)和(3,0).下列结论:①abc>0;②b-2a=0;③a-b+c<0;④a+b+c>0;⑤b2-4ac>0,其中正确的是( B )A.①②⑤B.①⑤C.②③D.①②③④⑤解析根据题图可得,抛物线开口向上,则a>0.抛物线与y轴的负半轴相交,则c<0,对称轴为直线x=-b2a ,−b2a>0,∴b<0,∴abc>0,故①正确;∵抛物线与x轴的两个交点分别为(-1,0),(3,0),∴对称轴是直线x=1,∴-b=1,2a∴b+2a=0,故②错误;当x=-1时,y=a-b+c=0,故③错误;由题图可知,当x=1时,y=a+b+c<0,故④错误;由题图知,抛物线与x轴有两个不同的交点,∴Δ=b2-4ac>0,故⑤正确.故选B.二、填空题(每小题3分,共15分)3+(-2 021)0= 0 .11.√-1解析原式=-1+1=0.12.如图,BD∥GE,AQ平分∠FAC,交BD于Q,∠GFA=50°,∠Q=25°,则∠ACB的度数为100°.解析如图,过点A作AH∥BD,∵BD∥GE,∴BD∥GE∥AH,∵∠GFA=50°,∠Q=25°,∴∠FAH=∠GFA=50°,∠HAQ=∠Q=25°,∴∠FAQ=∠FAH+∠HAQ=50°+25°=75°.∵AQ平分∠FAC,∴∠FAQ=∠CAQ=75°,∵∠CAQ+∠Q+∠ACQ=180°,∴∠ACQ=180°-75°-25°=80°,∴∠ACB=180°-∠ACQ=100°.13.如图,在△ABC中,AC=BC,∠B=70°,分别以点A、C为圆心,大于1AC的长为半径作弧,两弧相2交于点M、N,作直线MN,分别交AC、BC于点D、E,连接AE,则∠AED的度数是50°.解析由作图可知,直线MN是线段AC的垂直平分线,∴CE=AE,∠ADE=90°,∴∠C=∠CAE,∵AC=BC,∠B=70°,∴∠C=40°,∴∠CAE=40°,∴∠AED=50°.14.如图,在扇形AOB中,∠AOB=120°,连接AB,以OA为直径作半圆C,交AB于点D,若OA=6,.则图中阴影部分的面积为9π-27√34解析如图,连接OD、CD,∵OA为半圆C的直径,∴OD⊥AB,OC=AC,∵OA=OB=6,∠AOB=120°,∴AD=DB=1AB,∠OAD=30°,2OA=3,∴OD=12在Rt△ADO中,由勾股定理得,AD=√OA2-OD2=3√3,∴AB=2AD=6√3,∴S△AOB=1×AB×OD=9√3,2∵OC=AC,DB=AD,∴CD ∥OB,CD=12OB, ∴∠ACD=∠AOB=120°,S △ACD =14×S △AOB =9√34, ∴S 阴影=120π×62360-S △AOB -(120π×32360-S △ACD )=12π-9√3−3π+9√34=9π-27√34. 15.已知正方形ABCD 的边长为1,P 为AD 边上的动点(不与点A 重合),点A 关于直线BP 的对称点为E,连接PE,BE,CE,DE.当△CDE 是等腰三角形时,AP 的值为 2-√3或√33.解析 ①如图1,当CE=CD 时, 易得△BEC 是边长为1的等边三角形,过点E 作BC 的垂线,分别交AD,BC 于点M,N,则MN ⊥AD, 易得EN=√32BE =√32,∴ME=1-√32, 在四边形ABEP 中,∠ABE=30°,∠A=∠PEB=90°,∴∠APE=150°,∴∠MPE=180°-∠APE=30°,∴在Rt △PEM 中,PE=2ME=2-√3,∴AP=PE=2-√3;图1②如图2,当ED=CE时,点E在线段CD的垂直平分线上,也在线段AB的垂直平分线上,连接AE, ∴AE=BE,又∵AB=BE=1,∴△ABE是边长为1的等边三角形,∴∠ABE=60°,∴∠ABP=∠EBP=30°,∴在Rt△ABP中,AP=√33AB=√33.图2综上所述,AP的值为2-√3或√33.限时专练(二)(时间:22~25分钟 分值:45分)一、选择题(每小题3分,共30分) 1.12 021的相反数是( B ) A.12 021B.-12 021C.2 021 D .-2 021 解析12 021的相反数是-12 021,故选B.2.(2020洛阳三模)据统计,截至2020年1月末,我国外汇储备规模为31 155亿美元.将31 155亿用科学记数法表示为( C ) A.3.115 5×108 B .3.115 5×1011 C.3.115 5×1012D.3.115 5×1013解析 将31 155亿用科学记数法表示为3.115 5×1012.故选C.3.(2020贵州黔西南)如图是由6个相同的小正方体组合成的一个立体图形,它的俯视图为( D )解析从上面看可得四个并排的小正方形,如图所示:故选D.4.下列计算正确的是( D )A.a2·a3=a6B.-2(a-b)=-2a-2bC.2x2+3x2=5x4D.(-2a2)2=4a4解析 A.a2·a3=a5,故此选项错误;B.-2(a-b)=-2a+2b,故此选项错误;C.2x2+3x2=5x2,故此选项错误;D.(-2a2)2=4a4,故此选项正确.故选D.5.某学校准备从甲、乙、丙、丁四个小组中选出一个小组代表学校参加青少年科技创新大赛,各组平时成绩的平均数(分)及方差如表所示:甲乙丙丁x8 9 9 8s 2 1.1 1.3 1.1 1.4如果要选出一个成绩较好且状态稳定的组去参赛,那么应选择( B ) A.甲组 B.丙组 C.乙组 D.丁组解析 由题表可知,乙、丙的平均成绩较好,应从乙、丙中选择,由于s 乙2>s 丙2,故乙的方差大,波动大,则要选出一个成绩较好且状态稳定的组去参赛,应选择丙组. 故选B.6.化简分式a+1a 2-1的结果是( B ) A.aa -1 B.1a -1 C.1a+1 D.a+1 解析a+1a 2-1=a+1(a+1)(a -1)=1a -1,故选B.7.若关于y 的一元二次方程ky 2-2y-1=0有两个不相等的实数根,则k 的取值范围是( B ) A.k>-1 B.k>-1且k ≠0C.k<1D.k<1 且k≠0解析∵一元二次方程ky2-2y-1=0有两个不相等的实数根,∴Δ>0,即(-2)2-4k×(-1)>0,解得k>-1,又ky2-2y-1=0是关于y的一元二次方程,∴k≠0,∴k的取值范围是k>-1且k≠0,故选B.8.(2020湖北武汉)某班从甲、乙、丙、丁四位选手中随机选取两人参加校乒乓球比赛,恰好选中甲、乙两位选手的概率是( C )A.13B.14C.16D.18解析根据题意,画树状图如下:共有12种等可能的结果,其中恰好选中甲、乙两位选手的结果有2种,则恰好选中甲、乙两位选手的概率是212=16.故选C.9.(2020河南模拟)如图,动点P在平面直角坐标系中按图中箭头所示方向运动,第1次从原点运动到点(1,1),第2次接着运动到点(2,0),第3次接着运动到点(3,2),……,按这样的运动规律,经过第2 020次运动后,动点P的坐标是( B )A.(2 020,1)B.(2 020,0)C.(2 020,2)D.(2 019,0)解析动点P的坐标运动规律可以看作每运动4次一个循环,每个循环向右移动4个单位长度, 而2 020=505×4,所以,前505次循环运动点P共向右运动505×4=2 020个单位长度,且在x轴上,故经过第2 020次运动后,动点P的坐标为(2 020,0).故选B.10.(2020湖北铜仁)如图,在矩形ABCD中,AB=3,BC=4,动点P从点B开始沿B→C→D运动到点D,设点P运动的路程为x,△ADP的面积为y,那么y与x之间的函数关系的图象大致是( D )解析由题意得,AD=BC=4,AB=CD=3. 当0≤x≤4时,y=12×AD×AB=12×4×3=6.当4<x≤7时,y=12×PD×AD=12(7-x)×4=14-2x.故选D.二、填空题(每小题3分,共15分)11.(2020河南二模)计算:(-1)2-|√-83|= -1 .12.不等式组{x+5>3,x+6>4x-3的整数解是-1、0、1、2 . 解析解不等式x+5>3,得x>-2,解不等式x+6>4x-3,得x<3,则不等式组的解集为-2<x<3,所以不等式组的整数解为-1、0、1、2.13.(2020周口西华一模)已知抛物线y=-x2+bx+4经过(-3,m)和(5,m)两点,则b的值为 2 .解析由抛物线y=-x2+bx+4经过(-3,m)和(5,m)两点,可知抛物线的对称轴为直线x=1,∴-b2×(−1)=1,∴b=2.14.如图,在Rt△ABC中,∠ACB=90°,AC=BC=3,将Rt△ABC绕点A逆时针旋转30°后得到Rt△ADE,点B经过的路径为弧BD,则图中阴影部分的面积为3π2.解析∵∠ACB=90°,AC=BC=3,∴AB=3√2,又∵将Rt△ABC绕点A逆时针旋转30°后得到Rt△ADE,∴Rt△ADE≌Rt△ABC,S扇形BAD=30π×(3√2)2360=3π2,∴S阴影=S△ADE+S扇形BAD-S△ABC=S扇形BAD=3π2.15.如图,在矩形ABCD中,AB=2,AD=2√3,点E为线段AD的中点,动点F从点D出发,沿D→C→B的方向在DC和CB边上运动,将矩形沿EF折叠,点D的对应点为点D',当点D'恰好落在矩形.的对角线上时(不与矩形的顶点重合),点F运动的路程为1或2+√33解析如图1,当点D'落在对角线AC上时,连接DD',图1∵将矩形沿EF折叠,点D的对应点为点D',∴DD'⊥EF,∵点E为线段AD的中点,∴AE=ED=ED',∴∠EAD'=∠AD'E,∠ED'D=∠EDD',∵∠EAD'+∠AD'E+∠ED'D+∠EDD'=180°,∴∠AD'D=∠AD'E+∠ED'D=90°,即DD'⊥AC,∴EF∥AC,∴点F是CD的中点,∵在矩形ABCD中,AB=2,∴CD=AB=2,∴DF=1,∴点F运动的路程为1.如图2,当点D'落在对角线BD上时,作FH⊥AD于点H,易知四边形CFHD为矩形,图2在矩形ABCD中,AB=2,AD=2√3,∠C=∠ADC=90°,E为线段AD的中点,∴∠ADB=30°,ED=12AD=√3,∵EF⊥BD,∴∠FEH=60°,∵四边形CFHD为矩形,∴HF=CD=2,∴在Rt△EHF中,EH=HFtan60°=2√33,∵ED=√3,∴FC=HD=ED-EH=√33,∴点F运动的路程为2+√33.综上可知,点F运动的路程为1或2+√33.限时专练(三)(时间:22~25分钟分值:45分)一、选择题(每小题3分,共30分),2,-1这四个数中,最小的数是( D )1.在0,-13A.0B.-1C.2D.-13<0<2,解析∵-1<-13,2,-1这四个数中,最小的数是-1.故选D.∴在0,-132.近年来,全球发现了一种通过蚊虫进行传播的虫煤病毒——寨卡病毒,其直径约为0.000 002 1 cm.0.000 002 1用科学记数法表示为2.1×10n,则n为( A )A.-6B.-5C.5D.6解析0.000 002 1=2.1×10-6,故选A.3.将两个物体如图摆放,则它的俯视图是( A )解析圆柱体的俯视图是圆,正方体的俯视图是正方形,故选A.4.(2020河南模拟)解分式方程1+x-3x-2=1x-2时,去分母得( C )A.1+x+3=1B.x+2+x-3=1C.x-2+x-3=1D.x-2-x+3=15.某中学九年级二班的8名女同学在一次一分钟仰卧起坐测试中的成绩如下(单位:个):35;38;42;44;40;47;45;45,则这组数据的中位数、平均数分别是( B )A.42、42B.43、42C.43、43D.44、43解析把这组数据按从小到大排序,得35;38;40;42;44;45;45;47,则这组数据的中位数为42+442=43,平均数x=18×(35+38+42+44+40+47+45+45)=42,故选B.6.如图,AB∥CD,直线EF分别交AB、CD于点E、F,EG平分∠BEF,交CD于点G,∠1=40°,则∠2的度数是( B )A.60°B.70°C.55°D.40°解析∵AB∥CD,∴∠1+∠BEF=180°,∵∠1=40°,∴∠BEF=140°,∵EG平分∠BEF,∴∠BEG=12∠BEF=70°,∵AB∥CD,∴∠2=∠BEG=70°.故选B.7.若关于x的一元二次方程kx2-(2k+1)x+k+2=0有两个不相等的实数根,则k的取值范围是( D )A.k≤14B.k≤14且k≠0C.k<14D.k<14且k≠0解析∵关于x的一元二次方程kx2-(2k+1)x+k+2=0有两个不相等的实数根,∴{k≠0,Δ=[−(2k+1)]2-4k(k+2)>0,解得k<14且k≠0.故选D.8.如图,转盘中6个扇形的面积相等,任意转动转盘2次(当指针恰好指在分界线上时,重转),当转盘停止转动时,两次指针指向的数字之和是3的倍数的概率为( B )A.23B.13C.12D.25解析任意转动转盘2次,将所有结果列表如下:第二次第一次1 2 3 4 5 61 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) 其中,所有的结果数为36,两次指针指向的数字之和是3的倍数的结果数是12,∴两次指针指向的数字之和是3的倍数的概率P=1236=13.故选B.9.(2020湖北武汉)若点A(a-1,y1),B(a+1,y2)在反比例函数y=kx(k<0)的图象上,且y1>y2,则a的取值范围是( B )A.a<-1B.-1<a<1C.a>1D.a<-1或a>1解析∵k<0,∴在图象的每一支上,y随x的增大而增大,①当点A(a-1,y1),B(a+1,y2)在图象的同一支上时,∵y1>y2,∴a-1>a+1,此不等式无解;②当点A(a-1,y1),B(a+1,y2)在图象的两支上时,∵y1>y2,∴{a-1<0,a+1>0,解得-1<a<1.故选B.10.图1为一副重叠放置的三角板,其中∠ABC=∠EDF=90°,∠A=30°,∠E=45°,BC与DF共线,将△DEF沿CB方向平移,当EF经过AC的中点O时,如图2,直线EF交AB于点G,若BC=3,则此时OG的长度为( C )图1图2A.3B.32C.3√22D.3√32解析如图,过点O作OH⊥AG于点H,则∠AHO=∠OHG=90°,∵∠ABC=90°,∴∠FBG=90°,易得∠F=∠FGB=45°,∴∠OGA=45°,∵∠A=30°,BC=3,∴在Rt△ABC中,AC=2BC=6,∵点O是AC的中点,∴AO=12AC=3,∴在Rt△AHO中,OH=12AO=32,∴在Rt△OHG中,OG=√2OH=3√22,故选C.二、填空题(每小题3分,共15分) 11.计算:√9+√83= 5 . 解析 √9+√83=3+2=5.12.(2020开封二模)不等式组{1−x <0,13x −1≤0的解集是 1<x ≤3 .解析 解不等式1-x<0,得x>1, 解不等式13x-1≤0,得x ≤3, 则不等式组的解集为1<x ≤3.13.如图,在平行四边形ABCD 中,对角线AC,BD 相交于点O,AB=OB,点E,F 分别是OA,OD 的中点,连接EF,过点E 作EM ⊥BC 于点M,设EM 交BD 于点N,若∠CEF=45°,FN=5,则线段BC 的长为 4√5 .解析 设EF=x,∵点E,F 分别是OA,OD 的中点,∴EF是△OAD的中位线,∴AD=2x,AD∥EF,∴∠CAD=∠CEF=45°,∵四边形ABCD是平行四边形,∴AD∥BC,AD=BC=2x,∴∠ACB=∠CAD=45°,∵EM⊥BC,∴∠EMC=90°,∴△EMC是等腰直角三角形,∴∠CEM=45°,连接BE,∵AB=OB,点E是OA的中点,∴BE⊥OA于点E,∴∠BEM=45°, ∴BM=EM=MC=x,∴BM=FE,易得△ENF≌△MNB,∴EN=MN=12x,BN=FN=5,∴在Rt△MNB中,由勾股定理得BN2=BM2+MN2,即52=x2+(12x)2,解得x=2√5(负值舍去),∴BC=2x=4√5.14.(2020三门峡一模)如图,在扇形AOB中,∠AOB=90°,OB=2,DE是OA的垂直平分线,交OA于点D,交弧AB于点E,点C是OB的中点,连接AC,CE,则图中阴影部分的面积为2π3−12.解析如图,连接OE,过点E作EF⊥OB于点F,在扇形AOB中,∠AOB=90°,OB=2,点C是OB的中点,∴S扇形AOB=90×π×22360=π,OA=OB=2,OC=1,∴S△AOC=OC·OA2=1×22=1,∵DE是OA的垂直平分线,∴∠EDO=90°,OD=12OA=1,∵OE=OA=2,∴OD=12OE,∴∠DEO=30°,易得四边形DEFO是矩形,∴EF=OD=1,∠EOF=∠DEO=30°,∴S扇形EOB=30×π×22360=π3,∴S△OCE=OC·EF2=12,∴S阴影=S扇形AOB-S△AOC-(S扇形EOB-S△OCE)=π-1-π3+12=2π3−12.15.(2020商丘梁园模拟)如图,在矩形ABCD中,点P为AD边上的一个动点,以PB所在直线为对称轴将△APB折叠得到△EPB,点A的对称点为点E,射线BE交矩形ABCD的边于点 F,若AB=4,AD=6,则当点F为矩形ABCD的边的中点时,PA的长为43或4√10-43.解析当点F是AD边的中点时,如图1,图1 ∵四边形ABCD是矩形,∴∠A=90°,∵AD=6,∴AF=3,∴在Rt△BAF中,BF=√AB2+AF2=5,由折叠可知,AB=BE=4,∠PEF=90°,∴EF=5-4=1,设PA=PE=x,则PF=3-x, 在Rt △PEF 中,PE 2+EF 2=PF 2, 即x 2+12=(3-x)2,解得x=43,∴PA=43.当点F 是CD 边的中点时,如图2,延长AD 交射线BF 于点H,图2则CF=DF=2, ∵∠C=90°,BC=6, ∴BF=√BC 2+CF 2=2√10, ∵DH ∥BC,∴∠H=∠FBC, ∵∠DFH=∠BFC,DF=FC, ∴△DHF ≌△CBF(AAS), ∴DH=BC=6,FH=BF=2√10,∵AB=BE=4,∴EF=2√10−4,EH =2√10−4+2√10=4√10-4, 设PA=PE=y,则PD=6-y,PH=6-y+6=12-y, 易得∠PEH=90°,在Rt △PEH 中,PE 2+EH 2=PH 2, 即y 2+(4√10-4)2=(12-y)2, 解得y=4√10-43,∴PA=4√10-43. 综上所述,PA 的长为43或4√10-43. 限时专练(四)(时间:22~25分钟 分值:45分)一、选择题(每小题3分,共30分)1.在-5,0,-2,4这四个数中,最大的数是( A ) A.4B.-5C.0D.-2解析 根据有理数比较大小的法则,可得-5<-2<0<4,所以在-5,0,-2,4这四个数中,最大的数是4.故选A.2.(2020开封二模)被誉为“中国天眼”的世界上最大的单口径球面射电望远镜FAST,在理论上可以接收到137亿光年以外的电磁信号.数据137亿用科学记数法表示为( C )A.1.37×108B.1.37×109C.1.37×1010D.1.37×1011解析137亿=13 700 000 000=1.37×1010.故选C.3.下列图形中,既是轴对称图形,又是中心对称图形的是( A )解析 A.既是轴对称图形,又是中心对称图形,故此选项正确;B.不是轴对称图形,也不是中心对称图形,故此选项错误;C.是轴对称图形,不是中心对称图形,故此选项错误;D.不是轴对称图形,是中心对称图形,故此选项错误.故选A.4.方程x(x-4)+x-4=0的解是( C )A.x=4B.x=-4C.x=4或x=-1D.x=-1解析∵x(x-4)+x-4=0,∴(x-4)(x+1)=0,∴x-4=0或x+1=0,解得x=4或x=-1,故选C.的解集在数轴上表示正确的是( A )5.不等式组{x-1<1,x+1≥0解析解不等式x-1<1,得x<2,解不等式x+1≥0,得x≥-1,则不等式组的解集为-1≤x<2,故选A.6.(2020开封二模)如图,已知BM平分∠ABC,且BM∥AD,若∠ABC=70°,则∠A的度数是( B )A.30°B.35°C.40°D.70°解析∵BM平分∠ABC,∠ABC=70°,∠ABC=35°.∴∠MBA=12∵BM∥AD,∴∠A=∠MBA=35°.故选B.7.(2020河南模拟)如图,在△ABC中,以点A为圆心,AC的长为半径作弧,与BC交于点E,分别以点E,C为圆心,大于1EC的长为半径作弧,两弧相交于点P,作射线AP交BC于点D.若∠B=45°,∠2ACD=2∠CAD,则∠BAC的度数为( B )A.80°B.75°C.65°D.30°解析由作图过程可知,AP所在直线是线段EC的垂直平分线,∴∠ADB=∠ADC=90°,∵∠B=45°,∴∠BAD=45°,∵∠ACD=2∠CAD,∴3∠CAD=90°,∴∠CAD=30°,∴∠BAC=∠BAD+∠CAD=75°.故选B.8.(2019河南模拟)如图所示,小红制作了一个游戏转盘,红,绿两个区域扇形的圆心角度数分别为150°,90°.让转盘自由转动(落在边界处重转),指针停止后落在紫色区域的概率是( B )A.14B.13C.25D.512解析∵红,绿两个区域扇形的圆心角度数分别为150°,90°, ∴紫色区域扇形的圆心角度数为360°-150°-90°=120°,∴指针停止后落在紫色区域的概率是120360=13.故选B.9.(2020河南二模)如图,已知平行四边形ABCD的顶点A(-4,0),C(8,3),点B在x轴的正半轴上,点D 在y轴的正半轴上.连接AC,过点B作BE⊥CD,垂足为点E,BE交AC于点F,则点F的坐标为( D )A.(3,1)B.(4,1)C.(3,2)D.(4,2)解析设AC与OD交于点G. ∵四边形ABCD是平行四边形, ∴AB∥CD,AB=CD,∵AB⊥OD,∴CD⊥OD,∵A(-4,0),C(8,3),∴OA=4,AB=CD=8,OD=3,∴OB=AB-OA=4,∵AB∥CD,∴△OAG∽△DCG,∴OGDG =OACD=48=12,∴OG=12DG=13OD=1,∵BE⊥CD,CD⊥OD,∴OD∥BE,∴△AOG∽△ABF,∴OGBF =OAAB,即1BF=48,解得BF=2,∴点F的坐标为(4,2),故选D.10.(2020郑州二模)如图,在正方形ABCD中,CD=3 cm.动点P从点A出发,以√2 cm/s的速度沿AC方向运动到点C停止.动点Q同时从点A出发,以1 cm/s的速度沿A→B→C方向运动到点C 停止.设△APQ的面积为y(cm2),运动时间为x(s),则能反映y与x之间的函数关系的图象是( D )解析①当点Q在AB上运动时,过点P作PN⊥AB于点N,如图,y=12×AQ×PN=12x·√2x·sin∠BAC=12x·√2x·sin 45°=12x2(0≤x≤3);②当点Q在BC上运动时,y=12×AB ×QC =12×3×(6−x)=−32x+9(3<x ≤6). 故选D.二、填空题(每小题3分,共15分)11.(2020贵州遵义)计算√12−√3的结果是 √3 . 解析 √12−√3=2√3−√3=√3.12.返校复学前,小张进行了14天的体温测量,结果统计如下: 体温(℃) 36.3 36.4 36.5 36.6 36.7 36.8 天数123431则小张这14天体温的众数是 36.6 .解析 36.6出现的次数最多,有4次,所以众数是36.6.13.如果一组数据x 1,x 2,x 3,…,x n 的方差为3,那么另一组数据2x 1+2,2x 2+2,2x 3+2,…,2x n +2的方差为 12 .解析 ∵数据x 1,x 2,x 3,…,x n 的方差为3,∴数据2x 1+2,2x 2+2,2x 3+2,…,2x n +2的方差为22×3=12. 14.(2020河南模拟)如图,在Rt △ABC 中,点D 是AB 上的一点,将Rt △ABC 绕直角顶点C 逆时针旋转90°,使得点A 的对应点A'落在BC 的延长线上,点B 的对应点B'落在边AC 上,点D 的对应点D'落在边A'B'上,D'D ⏜经过点B',若AC=2BC=2√3,则图中阴影部分的面积是 9π4 .解析如图,连接CD、CD',由题意得,∠DCD'=∠ACA'=∠BCB'=90°, BC=CD=B'C=CD'=√3,AC=A′C=2√3, ∴∠BCD+∠DCB'=∠B'CD'+∠DCB'=90°, ∴∠BCD=∠B'CD',∴△DCB≌△D'CB'(SAS),由旋转可知,△ABC≌△A'B'C,∴S△DCB=S△D'CB',S△ABC=S△A'B'C,∴S△BCD+S△A'CD'=S△ABC,∴S阴影=S扇形ACA'+S△ABC-S扇形DCD'-S△BCD-S△A'CD' =S扇形ACA'+S△ABC-S扇形DCD'-(S△BCD+S△A'CD')=S扇形ACA'+S△ABC-S扇形DCD'-S△ABC =S扇形ACA'-S扇形DCD'=90×π×(2√3)2360−90×π×(√3)2360=9π4.15.如图所示,在正方形ABCD中,AB=8,BE=DF=1,M是射线AD上的动点,点A关于直线EM的对称点为A',当△A'FC是以FC为直角边的直角三角形时,MA的长为14√33或7√1313.解析如图1,当∠FCA'=90°,即点A'在BC上时,过点M作MN⊥BC于点N,则∠MNA'=90°,图1∵四边形ABCD是正方形,∴AB=CD=8,∠D=∠C=90°,∵MN⊥BC,∴四边形MNCD是矩形,∴MN=CD=8.∵AB=8,BE=DF=1,∴AE=CF=7.∵点A关于直线EM的对称点为A',∴AE=A'E=7,MA=MA',∠A=∠EA'M=90°. ∴在Rt△A'BE中,A'B=√A'E2-BE2=4√3. ∵∠BA'E+∠MA'N=90°,∠BA'E+∠A'EB=90°,∴∠A'EB=∠MA'N,∵∠B=∠MNA'=90°,∴△A'BE∽△MNA',∴A'BMN =A'EMA',即4√38=7MA',∴MA'=14√33=MA.如图2,当∠A'FC=90°时,过点A'作HG⊥AD,交AD于点H,交BC于点G,过点E作EK⊥HG于点K,∴∠EKA'=90°,∠MHA'=90°,图2∵四边形ABCD是正方形,∴AB=CD=8,∠D=∠DCB=90°,∵HG⊥AD, ∴四边形HGCD是矩形,∴HG=CD=8,同理可得,KG=BE=1,DF=A'H=1,AE=HK.∵AB=CD=8,BE=DF=1,∴AE=CF=7.∵点A关于直线EM的对称点为A',∴AE=A'E=7=HK,AM=A'M,∠A=∠EA'M=90°,∴A'K=HK-A'H=6,∴在Rt△EKA'中,EK=√A'E2-A'K2=√13.∵∠KA'E+∠MA'H=90°,∠KA'E+∠A'EK=90°,∴∠A'EK=∠MA'H,∵∠EKA'=∠MHA'=90°,∴△A'KE ∽△MHA',∴A'E MA'=EKA'H , 即7MA'=√131,∴MA'=7√1313=MA. 综上,MA 的长为14√33或7√1313.限时专练(五)(时间:22~25分钟 分值:45分)一、选择题(每小题3分,共30分) 1.(2020平顶山二模)比-2小的数是( A ) A.-4B.2C.-1D.3解析 A.-4<-2,故本选项符合题意; B.2>-2,故本选项不符合题意; C.-1>-2,故本选项不符合题意; D.3>-2,故本选项不符合题意. 故选A.2.(2020鹤壁一模)截至北京时间2020年4月11日21时,全球累计新冠肺炎确诊病例已超过171万.将171万用科学记数法表示为( C )A.1.71×105B.0.171×107C.1.71×106D.1 710 0003.(2020浙江台州)用三个相同的正方体搭成如图所示的立体图形,则该立体图形的主视图是( A )解析根据主视图的概念可知,选项A符合题意,故选A.4.(2020平顶山模拟)某花店连续六天销售玫瑰花的数量(单位:枝)分别为2,9,x,12,5,10,店主小明统计时发现,这6个数据的中位数与众数恰好相同,则x的值是( B )A.8B.9C.10D.115.如图,已知c⊥a,c⊥b,直线b,c,d交于一点,若∠1=48°,则∠2=( B )A.48°B.42°C.58°D.52°解析如图,∵c⊥a,c⊥b,∴∠4=∠5=90°,∴a∥b,∵∠1=48°,∴∠3=∠1=48°,∴∠2=180°-∠3-∠5=180°-48°-90°=42°, 故选B.6.程大位是我国明朝商人,珠算发明家,他60岁时完成的《直指算法统宗》是东方古代数学名著,详述了传统的珠算规则,确立了算盘用法,书中有如下问题:一百馒头一百僧,大僧三个更无争,小僧三人分一个,大小和尚得几丁.大意是:有100个和尚分100个馒头,如果大和尚1人分3个,小和尚3人分1个,正好分完,问大、小和尚各有多少人.设大和尚有x 人,小和尚有y 人,则下列方程组正确的是( A ) A.{x +y =1003x +y 3=100 B.{x +y =100x 3+y =100C.{x +y =1003x +y =100D.{x +y =100x +3y =1007.如图,在△ABC 中,∠C=150°,点E 是边AB 上一点,∠DEF=65°,则∠ADE+∠BFE=( D )A.180°B.185°C.205°D.215°解析在四边形CDEF中,∠C+∠CDE+∠CFE+∠DEF=360°,∠C=150°,∠DEF=65°,∴∠CDE+∠CFE=360°-65°-150°=145°,∴∠ADE+∠BFE=360°-(∠CDE+∠CFE)=215°,故选D.8.(2020开封二模)某校组织社团活动,小明和小刚从“数学社团”“航模社团”“文艺社团”三个社团中,各随机选择一个社团参加活动,两人恰好选择同一个社团的概率是( A )A.13B.2 3C.19D.29解析把“数学社团”“航模社团”“文艺社团”分别记为A、B、C,画树状图如下:共有9种等可能的结果,小明和小刚恰好选择同一个社团有3种结果,则小明和小刚恰好选择同一个社团的概率为39=13.故选A.9.如图,在Rt△ABC中,∠C=90°,以顶点A为圆心,适当长为半径画弧,分别交边AC、AB于点M、MN的长为半径画弧,两弧在∠CAB中交于点P,作射线AP交N,再分别以点M、N为圆心,大于12边BC于点D,若CD=4,AB=15,则△ABD的面积是( C )A.15B.45C.30D.60解析如图,作DE⊥AB于点E,由题意知,射线AP是∠CAB的平分线,∵∠C=90°,DE⊥AB,∴DE=CD=4,×AB×DE=30,∴△ABD的面积为12故选C.10.如图,在平面直角坐标系xOy中,矩形OABC的边OA在x轴上,OC在y轴上,OA=6,OC=4,点P 在边BC上,且PC=13BC.将矩形OABC绕点O以每秒45°的速度顺时针旋转,则第2 019秒时,点P的坐标为( C )A.(3√2,√2)B.(2,-1)C.(√2,−3√2)D.(-1,2)解析∵将矩形OABC绕点O以每秒45°的速度顺时针旋转,360°÷45°=8,∴每8秒循环一次,∵2 019÷8=252……3,∴第2 019秒时,矩形OABC旋转到图中矩形OA'B'C'的位置,作C'E⊥y轴于点E,P'F⊥C'E于点F,由题意可得△P'C'F,△OEC'都是等腰直角三角形,OC'=OC=4,PC=P'C'=13BC=13OA=2,∴OE=C'E=√22×4=2√2,P'F=C'F=√22×2=√2,∴EF=C'E-C'F=√2,∴点P'的坐标为(√2,−3√2),即第2 019秒时,点P的坐标为(√2,−3√2).故选C.二、填空题(每小题3分,共15分)3= π.11.计算:|3-π|-√-27解析原式=π-3+3=π.12.(2020开封一模)若关于x的一元二次方程2x2-4x=k没有实数根,则k的取值范围是k<-2 .解析原方程化为一般式为2x2-4x-k=0,根据题意,得Δ=(-4)2-4×2×(-k)<0,解得k<-2.(x<0)的图象经过点A,AB⊥x轴于点B,且△AOB的13.(2020湖南常德)如图,已知反比例函数y=kx面积为6,则k= -12 .。

2021年安徽中考数学总复习选择填空提分特训(03)

2021年安徽中考数学总复习选择填空提分特训(03)

选择填空提分特训(三)[限时:35分钟满分:60分]一、选择题(每小题4分,满分40分)1.-2的绝对值是()A.2B.-2C.12D.-122.计算-a×(-3a)2的结果是()A.-3a3B.6a2C.-9a3D.-3a23.据媒体报道,我国最新研制的“察打一体”无人机的速度极快,经测试最高速度可达204000米/分,这个数用科学记数法表示,正确的是()A.204×103B.20.4×104C.2.04×105D.2.04×1064.下面四个几何体中,俯视图是圆的几何体共有()图X3-1A.1个B.2个C.3个D.4个5.已知一元二次方程2x2-5x+3=0,则该方程根的情况是()A.有两个不相等的实数根B.有两个相等的实数根C.两个根都是自然数D.无实数根6.某超市老板购进一批大米,加价25%出售.为了为抗击疫情出一份力,该老板决定将该批大米降价至进价出售,则需要降()A.20%B.75%C.60%D.50%7.如图X3-2是某班测试1分钟跳绳成绩的统计图,成绩不低于140的为优秀等级,则该班跳绳优秀等级率为()图X3-2A.47.5%B.50%C.52.5%D.62.5%8.如图X3-3,△ABC是等腰直角三角形,AB=AC,点D是AC的中点.点E是BC上的一点,BD⊥DE,则CEBC的值为()图X3-3A.15B.16C.17D.2119.已知下列命题:①若a3>b3,则a2>b2;②若点A(x1,y1)和B(x2,y2)在二次函数y=x2-2x-1的图象上,且满足x1<x2<1,则y1>y2>-2;③在同一平面内,a,b,c是直线,且a∥b,b⊥c,则a∥c;④周长相等的所有等腰直角三角形全等.其中真命题的个数是()A.4个B.3个C.2个D.1个10.如图X3-4,在矩形ABCD中,△CEF是等腰直角三角形,且直角顶点E是AB上的点(点F在CE的左侧),若AB=8,BC=5,则AF的最小值为()图X3-4A.3√22B.√3C.2D.√39-22二、填空题(每小题5分,满分20分)11.因式分解:-2a2+2a-1=.212.如图X3-5,点A,B,C在☉O上,CO的延长线交AB于点D,∠A=50°,∠B=30°,则∠ADC的度数为.图X3-513.如图X3-6,反比例函数y=k(k≠0)的图象与正比例函数y=ax的图象交于点A,C,分别过点A,C作x轴的垂线,x垂足分别为点B,D.若四边形ABCD的面积为12,则k的值为.图X3-614.一副三角板如图X3-7放置,将三角板ADE绕点A逆时针旋转α(0°<α<90°),使得三角板ADE的一边所在的直线与BC垂直,则α的度数为.图X3-7附加训练15.如图X3-8是新农村广场的喷泉,其中有一股喷泉呈抛物线形状,如图②所示,抛物线与x轴,y轴交点分别为B(3,0),A(0,3),且抛物线的表达式满足函数关系y=-3x2+kx+b.2(1)求该抛物线的表达式;(2)求该股喷泉的最大高度;(3)在该抛物线上若有一点C,使得四边形AOBC的面积最大,求该点坐标.图X3-816.[原创题]如图①,△ABC和△DCE都是等腰直角三角形,AB=AC,DC=DE,且点A是DE上的点(异于两端点),过点B作BH∥AC交CE的延长线于点H,DE的延长线交BH于点G,过点A作AF∥CE交CD于点F,连接BE.(1)证明:△ABG∽△CDA;(2)若CH·CF=√2,求BC的长;(3)如图②,若△HEG≌△CEA,求tan∠CAF的值.图X3-9【参考答案】1.A2.C3.C4.B5.A6.A7.A8.B[解析]过点E作EF⊥AC于点F,则EF∥AB,则CEBC =EFAB.设EF=CF=a,AB=2b,则AD=CD=b,DF=b-a.易证△ABD∽△FDE,则EFAD =DFAB,即ab=b-a2b,化简,得3a=b,∴CEBC=EFAB=a2b=a6a=16.9.C[解析]若a3>b3,则a>b,但不能确定a2>b2,①为假命题;∵二次函数图象开口向上,对称轴为直线x=1,且满足x1<x2<1,在对称轴的左侧,y随x的增大而减小,但纵坐标大于最小值-2,②为真命题;在同一平面内,a,b,c是直线,且a∥b,b⊥c,则应该a⊥c,③为假命题;周长相等的所有等腰直角三角形全等,④为真命题.故选择C.10.A[解析]过点F作FG⊥AB于点G,在GB上截取GH=FG,连接FH,则△FGH是等腰直角三角形.∵∠FEG+∠BEC=∠ECB+∠BEC=90°,∴∠FEG=∠ECB.又∵EF=CE,∠FGE=∠CBE=90°,∴△EFG≌△CEB(AAS),∴EG=CB,FG=HG=BE,∴BH=EG=BC=5,即BH为定值.延长HF交AD于点I,则△AIH是等腰直角三角形,AH=AB-BH=3,∴IH=3√2,当AF⊥IH时,AF有最小值,最小值为3√22.11.-12(2a-1)212.110°13.-6[解析]由题意可知k<0,且点A,C关于原点对称,则OB=OD,OA=OC,故四边形ABCD是平行四边形,∴S△AOB=14S四边形ABCD=3,则k=-2S△AOB=-6.14.15°或60°[解析]分情况讨论:①当DE⊥BC时,∠BAD=75°,∴α=90°-∠BAD=15°;②当AD ⊥BC 时,∠BAD=30°,∴α=90°-∠BAD=60°. 故答案为:15°或60°.15.解:(1)将点A ,B 的坐标代入该函数表达式中,得: {b =3,-32×9+3k +b =0,解得:{k =72,b =3. 故该抛物线的表达式为y=-32x 2+72x+3.(2)由y=-32x 2+72x+3=-32x -762+12124可知: 当x=76时,该股喷泉的最大高度为12124米.(3)易知直线AB 的表达式为y=-x+3,设平行于AB 且经过点C 的直线解析式为y=-x+m ,联立抛物线和该直线的表达式,得:-32x 2+72x+3=-x+m ,即3x 2-9x+2m -6=0,当点C 距离直线AB 最大时,Δ=81-4×3(2m -6)=0,解得:m=518.解方程3x 2-9x+2×518-6=0,得:x 1=x 2=32,则y=-32+518=398.故当点C 的坐标为32,398时,四边形OACB 的面积最大.16.解:(1)证明:∵∠BAG+∠AGB=90°,∠BAG+∠CAD=180°-∠BAC=90°,∴∠AGB=∠CAD. 又∵∠ABG=∠CDA=90°, ∴△ABG ∽△CDA.(2)∵BH ∥AC ,CE ∥AF ,∴∠H=∠ACE=∠CAF .又∵∠CBH=∠ABH+∠ABC=135°,∠AFC=180°-∠AFD=135°,∴∠CBH=∠AFC ,∴△BCH ∽△FCA ,∴CHCA =CBCF,∴CH·CF=CA·CB=√2.又∵CA·CB=√22CB·CB=√2,∴BC=√2.(3)∵△HEG≌△CEA,∴AE=GE,即BE是Rt△ABG的中线,∴AE=BE.∵△ABG∽△CDA,∴∠BAE=∠ACF.由△ABC、△ADF、△DCE都是等腰直角三角形可知AB=AC,DE-AD=DC-DF,即AE=CF,∴△ABE≌△CAF,∴AF=BE=AE=CF.∴CD=DF+CF=DF+AF=DF+√2DF,∴tan∠CAF=tan∠ACD=ADCD =DF+2DF=√2-1.。

2021年山西中考数学总复习选择填空提分特训(03)

2021年山西中考数学总复习选择填空提分特训(03)

选择填空提分特训(三)一、选择题(每小题3分,共30分)1.√9的平方根是()A.3B.-3C.±3D.±√32.如图X3-1,已知直线AB,CD相交于点O,OA平分∠EOC,∠EOC=100°,则∠BOE的大小为()图X3-1A.100°B.110°C.120°D.130°3.小敏和小华在某次各科满分均为100分的期末测试中,各科成绩的平均分相同.小敏想和小华再比较一下两人中谁的各科成绩更加均衡,则她需要分别计算两人各科成绩的()A.加权平均数B.方差C.众数D.中位数4.若x,y的值均扩大为原来的2倍,则下列分式的值保持不变的是()A.xx-y B.2xy2C.x2yD.3x32y25.若a+b=5,ab=-3,则(a-b)2的值是()A.25B.19C.31D.376.如图X3-2,△ODC是由△OAB绕点O顺时针旋转30°后得到的图形.若点D恰好落在AB上,且∠AOC的度数为100°,则∠B的度数是()图X3-2A .40°B .35°C .30°D .15°7.若关于x 的不等式组{2x +7>4x +1,x -k <2的解集为x<3,则k 的取值范围为 ( )A .k>1B .k<1C .k ≥1D .k ≤18.如图X3-3,在Rt △ABC 中,∠C=90°,AB=6,AD 是∠BAC 的平分线,经过A ,D 两点的圆的圆心O 恰好落在AB 上,☉O 分别与AB ,AC 相交于点E ,F .若圆O 的半径为2,则阴影部分的面积为( )图X3-3A .13πB .43πC .23πD .92√3-3 9.根据学习函数的经验,小颖在平面直角坐标系中画出了函数y=4(x+2)2的图象,如图X3-4所示.根据图象,小颖得到了该函数的四条性质,其中正确的是( )图X3-4A .y 随x 的增大而增大B .当x>0时,0<y<1C .当x=-2时,y 有最大值D .当x=3与x=-3时,函数值相等10.如图X3-5,矩形OABC 的顶点O 与原点重合,点A ,C 分别在x 轴,y 轴上,点B 的坐标为(-5,4),点D 为边BC 上一动点,连接OD.若线段OD 绕点D 顺时针旋转90°后,点O 恰好落在AB 边上的点E 处,则点E 的坐标为( )图X3-5A .(-5,3)B .(-5,4)C .-5,52D .(-5,2)二、填空题(每小题3分,共15分)11.计算:√6×√23+√3= .12.某企业2018年获得利润300万元,计划2020年利润达到507万元,则这两年的年利润平均增长率为 .13.某篮球架的侧面示意图如图X3-6所示,现测得如下数据:底部支架AB 的长为1.74 m,后拉杆AE 的倾斜角∠EAB=53°,篮板MN 到立柱BC 的水平距离BH=1.74 m,在篮板MN 另一侧,与篮球架横伸臂DG 等高度处安装篮筐.已知篮筐到地面的距离GH 的标准高度为3.05 m,则篮球架横伸臂DG 的长约为 m .结果保留一位小数,参考数据:sin53°≈45,cos53°≈35,tan53°≈ 43图X3-614.如图X3-7,线段AB 的两个端点都在方格纸的格点上,建立平面直角坐标系后,A ,B 两点的坐标分别是(1,0)和(2,3).将线段AB 绕点A 逆时针旋转90°后再沿y 轴负方向平移4个单位,则此时点B 的对应点的坐标是 .图X3-715.如图X3-8,在△ABC中,∠BAC=90°,AC=6,BC=10,∠ABC和∠ACB的平分线相交于点O,过点O作OD∥AB 与边BC相交于点D,则OD的长为.图X3-8附加训练16.(1)计算:|-√2|+12-1-√6÷√3-2cos60°.(2)先化简,再求值:x2x-1-x2x2-1÷x2-xx2-2x+1,其中x=2.17.2020年4月22日是第51个世界地球日,地球日活动周中,同学们开展了丰富多彩的学习活动.某小组搜集到的数据显示,山西省总面积为15.66万平方公里,其中土石山区面积约为5.59万平方公里,其余部分为丘陵与平原,丘陵面积比平原面积的2倍还多0.8万平方公里.(1)求山西省的丘陵面积与平原面积;(2)活动周期间,两位家长计划带领若干学生去参观山西地质博物馆,他们联系了两家旅行社,报价均为每人30元.经协商,甲旅行社的优惠条件是,家长免费,学生都按九折收费;乙旅行社的优惠条件是,家长、学生都按八折收费.若只考虑收费,这两位家长应该选择哪家旅行社更合算?【参考答案】1.D2.D3.B4.A5.D6.B [解析]∵△COD 是△AOB 绕点O 顺时针旋转30°后得到的图形, ∴∠AOD=∠BOC=30°,AO=DO.∵∠AOC=100°,∴∠BOD=100°-30°×2=40°,∠ADO=∠A=12(180°-∠AOD )=12(180°-30°)=75°. 由三角形的外角性质,得∠B=∠ADO -∠BOD=75°-40°=35°.故选B . 7.C8.C [解析]连接OD ,OF .∵AD 是∠BAC 的平分线, ∴∠DAB=∠DAC.∵OD=OA ,∴∠ODA=∠OAD ,∴∠ODA=∠DAC , ∴OD ∥AC ,∴∠ODB=∠C=90°,S △AFD =S △OF A , ∴S 阴=S 扇形OF A .∵OD=OA=2,AB=6,∴OB=4,∴OB=2OD , ∴∠B=30°,∴∠BAC=60°.∵OF=OA ,∴△AOF 是等边三角形,∴∠AOF=60°,∴S 阴=S 扇形OF A =60π·22360=2π3.故选C .9.B10.A [解析]由题可得AO=BC=5,AB=CO=4.由旋转可得DE=OD ,∠EDO=90°,又∵∠B=∠OCD=90°,∴∠EDB+∠CDO=90°=∠COD+∠CDO.∴∠EDB=∠DOC.∴△DBE≌△OCD(AAS).∴BD=OC=4.设AE=x,则BE=4-x=CD.∵BD+CD=5,∴4+4-x=5.解得x=3.∴AE=3.∴E(-5,3).故选A.11.√312.30%13.1.2[解析]如图,延长AC,HG,交于点O.由题意,得DG∥AB,∴△OAH∽△ODG.∴OHAH =OGDG=tan53°≈43.∵AH=AB+BH=3.48(m),∴OH≈43AH=4.64(m).∵GH=3.05 m,∴OG=OH-GH=1.59(m).∴DG≈34OG≈1.2(m).14.(-2,-3)15.103[解析]由勾股定理得AB=8.作OE∥AC交BC于点E,则∠OED=∠ACB.∵OD∥AB,∴∠ODE=∠ABC,∴∠DOE=90°,∴△DOE∽△BAC,∴OE∶OD∶DE=AC∶AB∶BC=6∶8∶10.∵OD∥AB,∴∠ABO=∠BOD.∵∠ABO=∠OBD ,∴∠BOD=∠OBD ,∴BD=OD , 同理CE=OE ,∵BD+DE+EC=BC=10,∴OD+DE+OE=10, 又OD ∶DE ∶OE=8∶10∶6, ∴OD=10×824=103.16.解:(1)原式=√2+2-√6÷3-2×12 =√2+2-√2-1=1.(2)原式=x 3+x 2(x+1)(x -1)-x 2(x+1)(x -1)·(x -1)2x (x -1)=x 3(x+1)(x -1)·(x -1)2x (x -1)=x 2x+1.当x=2时,原式=43.17.解:(1)设山西省的平原面积为x 万平方公里,则山西省的丘陵面积为(2x+0.8)万平方公里. 由题意得x+2x+0.8+5.59=15.66,解得x=3.09,2x+0.8=6.98.答:山西省的平原面积为3.09万平方公里,山西省的丘陵面积为6.98万平方公里. (2)设去参观山西地质博物馆的学生有m 人,甲、乙旅行社的收费分别为y 甲元,y 乙元. 由题意得y 甲=30×0.9m=27m ,y 乙=30×0.8(m+2)=24m+48,当y 甲=y 乙时,27m=24m+48,m=16;当y 甲>y 乙时,27m>24m+48,m>16;当y 甲<y 乙时,27m<24m+48,m<16.答:当学生人数为16人时,两家旅行社的费用一样;当学生人数大于16人时,选择乙旅行社比较合算;当学生人数小于16人时,选择甲旅行社比较合算.。

2021年浙江中考数学总复习选择填空提分特训(03)

2021年浙江中考数学总复习选择填空提分特训(03)

选择填空提分特训(三)一、选择题(每小题3分,共30分)1.下列各组数中,不是互为相反数的是()A.-(-3)与+(-3)B.-32与(-3)2C.-|-3|与|+3|D.-(-3)3与332.熔喷布,俗称口罩的“心脏”,是口罩中间的过滤层,能过滤细菌,阻止病菌传播.经测量,医用外科口罩的熔喷布厚度约为0.000156米,将0.000156用科学记数法表示应为()A.0.156×10-3B.1.56×10-3C.1.56×10-4D.15.6×10-43.如图X3-1所示的几何体的左视图是()图X3-1图X3-24.在一个不透明的布袋中装有6个只有颜色不同的小球,其中3个红球、2个白球和1个黄球,从袋中任意摸出一个球,是白球的概率为()A.16B.13C.12D.565.下列计算正确的是()A.3a+2b=5abB.(a3)2=a6C.a6÷a3=a2D.(a+b)2=a2+b26.每年的3月12日是我国的植树节,某学校在“爱护生命,绿化祖国”的活动中,组织了100名学生开展植树造林活动,其植树情况整理如表,则这100名学生所植树的中位数和众数分别为()植树棵数45679人数302027158A.4,6B.5,9C.5.5,4D.6,97.将抛物线y=x2-4x+3平移,使它平移后的顶点坐标为(-2,4),则可将该抛物线()A.先向右平移4个单位,再向上平移5个单位B.先向右平移4个单位,再向下平移5个单位C.先向左平移4个单位,再向上平移5个单位D.先向左平移4个单位,再向下平移5个单位8.如图X3-3,AB是☉O的弦,AC是☉O的切线,A为切点,BC经过圆心.若∠B=20°,则∠C的度数为()图X3-3A.20°B.25°C.40°D.50°9.如图X3-4,正方形ABCD的边长为6,点E,F分别为AB,BC的中点,把△DCF沿DF折叠得到△DC'F,延长DC'交AB于点G,连结FG,H是AD边上的一点,连结EH,把△AEH沿EH折叠,点A的对应点A'恰好落在DC'上,则下列结论错误的是()图X3-4A .△FBG ≌△FG'GB .∠DFG=90°C .sin ∠AGD=45D .A'G=3.610.如图X3-5a,现有8枚棋子呈一直线摆放,依次编号为①~⑧.小明进行隔子跳,想把它跳成4叠,每2枚棋子为一叠.隔子跳规则为:只能靠跳跃,每一步跳跃只能是把一枚棋子跳过两枚棋子与另一枚棋子相叠,如图b 中的(1)或(2)(可随意选择跳跃方向),一枚棋子最多只能跳一次.若小明只通过4步便跳跃成功,那么他的第一步跳跃可以为( )图X3-5A .①叠到④上面B .②叠到⑤上面C .④叠到⑦上面D .⑤叠到⑧上面二、填空题(每小题4分,共24分) 11.因式分解:x 3-x= . 12.化简:2aa 2-4-1a -2= .13.我国古代数学著作《算法统宗》里有这样一首诗:我问开店李三公,众客都来到店中,一房七客多七客,一房九客一房空.问几房几客?意思是:“一批客人来到李三店中住宿,如果每一间客房住7人,那么有7人无房可住;如果每一间客房住9人,那么就空出1间房.问有多少房间,多少客人?”那么房间有 间,客人有 人.14.如图X3-6,在矩形ABCD 中,E 是直线BC 上一点,且CE=CA ,连结AE.若∠BAC=60°,则∠CAE 的度数为 .图X3-615.如图X3-7,过点C (4,5)的直线y=53x+b 交x 轴于点A ,∠ABC=90°,AB=CB ,反比例函数y=kx(x>0)的图象过点B ,将点C 沿x 轴的负方向平移a 个单位长度后恰好落在该反比例函数的图象上,则a 的值为 .图X3-716.如图X3-8,在菱形ABCD 中,∠B=60°,AB=2,若点P 是菱形ABCD 内部或边上一点,满足△PBC 是等腰三角形,则P ,D (P ,D 不重合)两点间距离的最小值为 .图X3-8附加训练17.计算:|√3-2|+20200--13-1+3tan30°+√8.18.解方程:xx -1+21-x=3.19.先化简,再求值:2b2+(a+b)(a-b)-(a-b)2,其中a=-3,b=1.2【参考答案】1.D2.C3.D4.B5.B6.C7.C8.D9.D [解析] 在Rt △FBG 和Rt △FC'G 中,{BF =FC =FC ',FG =FG ,∴Rt △FBG ≌Rt △FC'G ,故A 正确; ∵Rt △FBG ≌Rt △FC'G ,∴∠BFG =∠C'FG.又∵∠CFD =∠DFC',∴∠DFG =∠GFC'+∠DFC'=90°,故B 正确; 设BG =x ,则GC'=x ,DG =DC'+GC'=6+x. 在Rt △DFG 中,DG 2=GF 2+FD 2, ∴(x +6)2=(x 2+9)+(9+36),解得x =32,∴DG =GC'+6=152,sin ∠AGD =ADDG =45,故C 正确; 作EM ⊥DG 于点M ,则sin ∠EGM =sin ∠AGD =45,即EM EG=45,可得EM =65,∴A'M =√A 'E 2-EM 2=√32-(65) 2=3√215, GM =√EG 22=√(32) 2-(65) 2=910,∴A'G =A'M +MG =3√215+910≠3.6.故D 错误.10.C11.x (x +1)(x -1) 12.1a+213.8 63 14.15°或75°15.3 [解析] 作CD ⊥x 轴于点D ,BF ⊥x 轴于点F ,过B 作BE ⊥CD 于点E ,∵点C (4,5)在直线y =53x +b 上,∴5=4×53+b ,解得b =-53,∴直线解析式为y =53x -53,令y =0,则求得x =1,∴A (1,0).∵BF ⊥x 轴,BE ⊥CD ,∴BE ∥x 轴,∴∠ABE =∠BAF . ∵∠ABC =90°,∴∠ABE +∠EBC =90°. ∵∠BAF +∠ABF =90°,∴∠EBC =∠ABF .在△EBC 和△FBA 中,∠EBC =∠FBA ,∠BEC =∠BF A =90°,BC =AB , ∴△EBC ≌△FBA (AAS ),∴CE =AF ,BE =BF .设B c ,k c ,则5-k c =c -1,c -4=kc ,解得c =5.过点C 作x 轴的平行线交反比例函数图象于点G ,由B 5,k5,E 4,k5,BF =BE =1=k5,得k =5, ∴G (4-a ,5),即5=k4-a,解得a =3.16.2√3-2 [解析] ①若以边BC 为底,则BC 垂直平分线上(在菱形的边及其内部)的点满足题意,此时当点P 与点A 重合时,PD 的值最小,为2;②若以边PC 为底,∠PBC 为顶角时,以点B 为圆心,BC 长为半径作圆,与BD 相交于一点,则AC ⏜(除点C 外)上的所有点都满足△PBC 是等腰三角形,当点P 在BD 上时,PD 最小,最小值为2√3-2;③若以边PB 为底,∠PCB 为顶角,以点C 为圆心,BC 为半径作圆,则BD ⏜上的点A ,点D 均满足△PBC 为等腰三角形,当点P 与点D 重合时,显然不满足题意,故此种情况不存在. 综上所述,P ,D (P ,D 不重合)两点间距离的最小值为2√3-2.17.解:原式=2-√3+1-(-3)+3×√33+2√2=6+2√2. 18.解:去分母得x +(-2)=3(x -1),∴2x =1,解得x =12.经检验,x =12是原方程的解, ∴原方程的解为x =12.19.解:原式=2b 2+a 2-b 2-(a 2-2ab +b 2)=a 2+b 2-a 2+2ab -b 2=2ab. ∵a =-3,b =12,∴原式=2×(-3)×12=-3.。

2021年中考化学一轮复习限时训练3 我们周围的空气

2021年中考化学一轮复习限时训练3 我们周围的空气

限时训练3我们周围的空气一、单项选择题1．(2020兰州中考)下列用于食品包装中防腐的气体是()A．氧气 B．一氧化氮C．二氧化碳 D．氮气2．下列物质中，目前被计入“空气污染指数”项目的是()A．氮气 B．氧气C．稀有气体 D．可吸入颗粒物3．霓虹灯让我们的生活亮丽多彩。

霓虹灯中填充的气体是()A．氧气 B．氮气C．稀有气体 D．二氧化碳4．(2020长沙中考)今年，长沙的空气质量明显改善，空气是一种宝贵的自然资源。

下列说法正确的是() A．洁净的空气是纯净物B．空气中体积分数最大的是氮气C．氧气具有助燃性，常用作燃料D．空气中臭氧(O3)含量少，属于稀有气体5．下列实验现象的描述中，正确的是()A．铁丝在氧气中燃烧，火星四射，生成黑色固体B．硫在氧气中燃烧，发出淡蓝色火焰C．红磷在空气中燃烧产生大量的白雾D．木炭在氧气中燃烧，生成有刺激性气味的气体6．(2020铜仁中考)生活中的下列物质属于纯净物的是()A．酱油 B．豆浆 C．海水 D．氧气7．有三瓶无色、无味的气体，分别是氧气、空气、二氧化碳，区别它们最简单的方法是()A．分别测定气体的密度B．分别伸入燃着的木条C．分别滴入澄清石灰水，振荡D．分别滴入紫色石蕊溶液，振荡8．空气是人类宝贵的自然资源。

下列关于空气的说法错误的是()A．空气的主要成分是氮气和氧气B.造成空气污染的气体主要有CO、CO2和SO2等C．空气中的氧气可以供给呼吸、支持燃烧D．工业上常用分离液态空气的方法制备氧气9．对一定量氯酸钾和二氧化锰的混合物加热，下列图象能正确表示对应变化关系的是()10．(2020北京中考)利用如图装置验证了空气中氧气的含量。

下列叙述不正确的是()A．实验前需检查装置气密性B．白磷的作用是消耗试管中的氧气C．白磷熄灭、试管冷却后再打开止水夹D．最终量筒中液面约降至40 mL刻度线处二、填空题11．“雾霾”对人体健康有显著的负面影响。

(1)“雾霾”主要是由直径小于或等于__2__μm的可吸入颗粒物等排放产生的。

浙江省2021年中考数学复习选择填空限时练习汇总

浙江省2021年中考数学复习选择填空限时练习汇总

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通过我们的努力，能够为您解决问题，这是我们的宗旨，欢迎您下载使用!超级资源（共９套45页）浙江省2019年中考数学复习选择填空限时练习汇总选择填空限时练(一)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.-2的相反数是( )A. B.-2 C.2 D.-2.如图X1-1,下面几何体的俯视图是( )图X1-1图X1-23.[2018·绍兴]绿水青山就是金山银山,为了创造良好的生态生活环境,浙江省2017年清理河湖库塘淤泥约116000000方,数字116000000用科学记数法可以表示为 ( )A.1.16×109B.1.16×108C.1.16×107D.0.116×1094.把不等式组的解表示在数轴上,下列选项正确的是( )图X1-35.用反证法证明命题:在一个三角形中,至少有一个内角不大于60°.证明的第一步是( )A.假设三个内角都不大于60°B.假设三个内角都大于60°C.假设三个内角至多有一个大于60°D.假设三个内角至多有两个大于60°6.从某市8所学校中抽取共1000名学生进行800米跑达标抽样检测,结果显示该市成绩达标的学生人数超过半数,达标率达到52.5%.如图X1-4①、②反映的是本次抽样中的具体数据. 根据数据信息,下列判断:①小学高年级被抽检人数为200人;②小学、初中、高中学生中,高中生800米跑达标率最大;③小学生800米跑达标率低于33%;④高中生800米跑达标率超过70%.其中判断正确的有( )图X1-4A.0个B.1个C.2个D.3个7.如图X1-5,在▱ABCD中,用直尺和圆规作∠BAD的平分线AG交BC于点E,若BF=6,AB=5,则AE的长为( )图X1-5A.4B.6C.8D.108.已知关于x的方程ax+b=0(a≠0)的解为x=-2,点(1,3)是抛物线y=ax2+bx+c(a≠0)上的一个点,则下列四个点中一定在该抛物线上的是 ( )图X1-6A.(2,3)B.(0,3)C.(-1,3)D.(-3,3)9.如图X1-6,已知A,B是反比例函数y=(k>0,x>0)图象上的两点,BC∥x轴,交y轴于点C,动点P从坐标原点O出发,沿O→A→B→C匀速运动,终点为C,过运动路线上任意一点P作PM ⊥x轴于点M,PN⊥y轴于点N,设四边形OMPN的面积为S,P点运动的时间为t,则S关于t的函数图象大致是( )图X1-710.如图X1-8,正方形ABCD的边长为6,点E,F分别在AB,AD上,若CE=3,且∠ECF=45°,则CF的长为 ( )图X1-8A.2B.3C. D.二、填空题(每小题4分,共24分)11.一组数据2,3,3,5,7的中位数是,方差是.12.如图X1-9是一个斜体的“土”字,AB∥CD,已知∠1=75°,则∠2= °.图X1-913.为了了解某毕业班学生的睡眠时间情况,小红随机调查了该班15名同学,结果如下表:每天睡眠时间7 7.5 8 8.5 9 (单位:小时)人数 2 4 5 3 1 则这15名同学每天睡眠时间的众数是小时,中位数是小时.14.如图X1-10,将弧长为6π的扇形纸片AOB围成圆锥形纸帽,使扇形的两条半径OA与OB重合(粘连部分忽略不计),则圆锥形纸帽的底面圆半径是.图X1-1015.如图X1-11,已知点B,D在反比例函数y=(a>0)的图象上,点A,C在反比例函数y=(b<0)的图象上,AB∥CD∥x轴,AB,CD在x轴的同侧,AB=4,CD=3,AB与CD间的距离为1,则a-b的值是.图X1-1116.如图X1-12,点A(2,0),以OA为半径在第一象限内作圆弧AB,使∠AOB=60°,点C为弧AB 的中点,D为半径OA上一动点(不与点O,A重合),点A关于直线CD的对称点为E,若点E落在半径OA上,则点E的坐标为;若点E落在半径OB上,则点E的坐标为.图X1-12|加加练|1.计算:+20170-(-)-1+3tan30°+.2.解方程:+=3.3.先化简,再求值:2b2+(a+b)(a-b)-(a-b)2,其中a=-3,b=.参考答案1.C2.A3.B4.B5.B6.C7.C8.D9.B10.A11.312.10513.8814.315.1216.(2-2,0)(-1,3-)加加练1.解:原式=2-+1-(-3)+3×+2=6+2.2.解:去分母得x+(-2)=3(x-1),∴2x=1,∴x=.经检验,x=是原方程的解,∴原方程的解为x=.3.解:原式=2b2+a2-b2-(a2-2ab+b2)=a2+b2-a2+2ab-b2=2ab.∵a=-3,b=,∴原式=2×(-3)×=-3.选择填空限时练(二)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.某小区经过改进用水设施,5年内小区居民累计节水39400吨,将39400用科学记数法表示为( )A.3.9×104B.3.94×104C.39.4×103D.4.0×1042.下列运算正确的是( )A.(-3)2=-9B.(-1)2015×1=-1C.-5+3=8D.-|-2|=23.下列图形中是轴对称图形但不是中心对称图形的是( )A.等边三角形B.平行四边形C.矩形D.圆4.不等式3x<2(x+2)的解是( )A.x>2B.x<2C.x>4D.x<45.已知一组数据0,-1,1,2,3,则这组数据的方差为( )A.0B.1C.D.26.在Rt△ABC中,两直角边的长分别为6和8,则其斜边上的中线长为( )A.10B.3C.4D.57.在☉O中,圆心O到弦AB的距离为AB长度的一半,则弦AB所对圆心角的大小为( )A.30°B.45°C.60°D.90°8.已知点C是线段AB的黄金分割点(AC>BC),则下列结论正确的是( )A.AB2=AC2+BC2B.BC2=AC·BAC.=D.=9.如图X2-1,D是等边三角形ABC边AB上的一点,且AD∶DB=1∶2,现将△ABC折叠,使点C 与D重合,折痕为EF,点E,F分别在AC和BC上,则CE∶CF= ( )图X2-1A. B. C. D.10.若二次函数y=ax2+bx+c(a≠0)的图象与x轴的交点坐标分别为(x1,0),(x2,0),且x1<x2,图象上有一点M(x0,y0)在x轴下方,对于以下说法:①b2-4ac>0;②x=x0是方程ax2+bx+c=y0的解;③x1<x0<x2;④a(x0-x1)(x0-x2)<0.其中正确的是( )A.①③④B.①②④C.①②③D.②③二、填空题(每小题4分,共24分)11.请写出一个解为x=1的一元一次方程: .12.计算:2tan60°+(2-)0-()-1= .13.二次函数y=x2+4x+5(-3≤x≤0)的最大值是,最小值是.14.当1<a<2时,代数式+|1-a|= .15.如图X2-2,已知点A1,A2,…,A n均在直线y=x-1上,点B1,B2,…,B n均在双曲线y=-上,并且满足:A1B1⊥x轴,B1A2⊥y轴,A2B2⊥x轴,B2A3⊥y轴,…,A n B n⊥x轴,B n A n+1⊥y轴,…,记点A n的横坐标为a n(n为正整数).若a1=-1,则a3= ,a2015= .图X2-216.如图X2-3,在边长为2的菱形ABCD中,∠A=60°,M是AD边的中点,N是AB边上一动点,将△AMN沿MN所在的直线翻折得到△A'MN,连结A'C,则A'C长度的最小值是.图X2-3参考答案1.B2.B3.A4.D5.D6.D7.D8.C9.A10.B11.x-1=0(答案不唯一)12.2-113.5 114.115.216.-1选择填空限时练(三)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.-5的绝对值等于( )A.5B.-5C. D.-2.下列几何体中,俯视图为三角形的是( )图X3-13.事件:在只装有2个红球和8个黑球的袋子里,摸出一个白球是( )A.可能事件B.随机事件C.不可能事件D.必然事件4.下列运算正确的是( )A.(2a2)3=6a6B.-a2b2·3ab3=-3a2b5C.+=-1D.·=-15.在一次中学生田径运动会上,参加男子跳高的20名运动员的成绩如下表:成绩/米 1.55 1.60 1.65 1.70 1.75 1.80 人数 4 3 5 6 1 1 则这些运动员成绩的众数为( )A.1.55米B.1.65米C.1.70米D.1.80米6.已知点(-2,y1),(3,y2)在一次函数y=2x-3的图象上,则y1,y2,0的大小关系是( )A.y1<y2<0B.y1<0<y2C.y2<0<y1D.0<y1<y27.如图X3-2,一架长2.5米的梯子AB斜靠在墙上,已知梯子底端B到墙角C的距离为1.5米,设梯子与地面所夹的锐角为α,则cosα的值为( )图X3-2A. B. C. D.8.我们知道方程组的解是现给出另一个方程组它的解是( )A. B.C. D.9.七巧板是我们祖先的一项卓越创造,被誉为“东方魔板”.如图X3-3是一个七巧板迷宫,它恰好拼成了一个正方形ABCD,其中E,P分别是AD,CD的中点,一只蚂蚁从点A处沿图中实线爬行到出口点P处.若AB=2,则它爬行的最短路程为( )图X3-3A. B.1+C.2D.310.如图X3-4,在▱ABCD中,∠DAB=60°,AB=10,AD=6,☉O分别切边AB,AD于点E,F,且圆心O 恰好落在DE上.现将☉O沿AB方向滚动到与边BC相切(点O在▱ABCD的内部),则圆心O移动的路径长为( )图X3-4A.4B.6C.7-D.10-2二、填空题(每小题4分,共24分)11.分解因式:ab+ac= .12.小红同学5月份各项消费情况的扇形统计图如图X3-5,其中小红在学习用品上支出100元,则在午餐上支出元.图X3-513.如图X3-6,在☉O中,C为优弧AB上一点,若∠ACB=40°,则∠AOB= 度.图X3-614.甲、乙两工程队分别承接了250米,150米的道路铺设任务,已知乙比甲每天多铺设5米,甲完成铺设任务的时间是乙的2倍.设甲每天铺设x米,则根据题意可列出方程: .15.如图X3-7,点A在第一象限,作AB⊥x轴,垂足为点B,反比例函数y=的图象经过AB的中点C,过点A作AD∥x轴,交该函数图象于点D.E是AC的中点,连结OE,将△OBE沿直线OE对折到△OB'E,使OB'恰好经过点D,若B'D=AE=1,则k的值是.图X3-716.如图X3-8,矩形ABCD和正方形EFGH的中心重合,AB=12,BC=16,EF=,分别延长FE,GF,HG和EH交AB,BC,CD,AD于点I,J,K,L.若tan∠ALE=3,则AI的长为,四边形AIEL的面积为.图X3-8|加加练|1.计算:(-2018)0+-9×.2.化简:(a+2)(a-2)-a(a+1).3.化简:+.参考答案1.A2.C3.C4.C5.C6.B7.A8.D9.B[解析] ∵正方形ABCD,E,P分别是AD,CD的中点,AB=2,∴AE=DE=DP=1,∠D=90°,∴EP==,∴蚂蚁从点A沿图中实线爬到出口点P处,爬行的最短路程为AE+EP=1+.故选B.10.B[解析] 连结OA,OF.∵AB,AD分别与☉O相切于点E,F,∴OE⊥AB,OF⊥AD,∴∠OAE=∠OAD=30°.在Rt△ADE中,AD=6,∠ADE=30°,∴AE=AD=3,∴OE=AE·=.∵AD∥BC,∠DAB=60°,∴∠ABC=120°.设当运动停止时,☉O与BC,AB分别相切于点M,N,连结ON,OM,OB.则∠BON=30°,且ON=,∴BN=ON·tan 30°=1,EN=AB-AE-BN=10-3-1=6.∴圆心O移动的路径长为6.11.a(b+c)12.20013.8014.=15.12[解析] 如图,过D作DF⊥OB于F,设B'E与AD交于点G.∵AB⊥x轴,AD∥x轴,∴四边形ABFD是矩形,由折叠可得,∠B'=90°=∠A.又∵B'D=AE=1,∠DGB'=∠EGA,∴△DB'G≌△EAG,∴DG=EG,B'G=AG,∴AD=B'E=BE.又∵E是AC的中点,C是AB的中点,∴AE=CE=1,AC=BC=2,∴BE=3=AD,AB=4=DF.设C(a,2),则D(a-3,4).∵反比例函数y=的图象经过点C,D,∴2a=4(a-3),解得a=6,∴C(6,2),∴k=6×2=12.16.5[解析] 如图,过点E作EM⊥AB于点M,过点F作FN⊥AB于点N,过点E作EA1⊥AD于点A1,交FN于Q,过点G作GA2⊥AD,过点H作HP⊥A1E于P,∵tan∠1=3,∴tan∠2=3.又∵EF=,∴EQ=1,QF=3.∵矩形ABCD与正方形EFGH的中心重合,∴AA1=A2D=6,A1A2=4=PQ.同理得AN=8,NB=4,EM=6.易证△IME∽△EQF,∴=,∴IM=2,∴IB=7,∴AI=5.∴A1E=7,∴A1L=,∴四边形AIEL的面积为+=×(5+7)×6+×7×=.加加练1.解:原式=1+2-9×=2.2.解:原式=a2-4-a2-a=-4-a.3.解:原式===a.选择填空限时练(四)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.下列四个数:-1,0,,3.14,其中为无理数的是( )A.-1B.0C.D.3.142.下列计算正确的是( )A.x3+x4=x7B.x3-x4=x-1C.x3·x4=x7D.x3÷x4=x3.如图X4-1所示的支架的主视图是 ( )图X4-1图X4-24.如图X4-3,电路图上有四个开关A,B,C,D和一个小灯泡,闭合开关D或同时闭合开关A,B,C 都可使小灯泡发光,则任意闭合其中两个开关,小灯泡发光的概率是( )图X4-3A. B.C. D.5.如图X4-4,已知直线AB∥CD,∠GEB的平分线EF交CD于点F,∠1=60°,则∠2等于( )图X4-4A.130°B.140°C.150°D.160°6.若a-b=2ab,则-的值为 ( )A.-2B.-C.D.27.若将直尺的0 cm刻度线与半径为5 cm的量角器的0°线对齐,并让量角器沿直尺的边缘无滑动地滚动(如图X4-5),则直尺上的10 cm刻度线对应量角器上的度数约为( )图X4-5A.90°B.115°C.125°D.180°8.在某次体育测试中,九年级一班女同学的一分钟仰卧起坐成绩(单位:个)如下表:成绩45 46 47 48 49 50 人数 1 2 4 2 5 1 这次测试成绩的中位数和众数分别为( )A.47,49B.48,49C.47.5,49D.48,509.如图X4-6,在矩形ABCD中,AB=3,BC=5,点P是BC边上的一个动点(点P不与点B,C重合),现将△PCD沿直线PD折叠,使点C落到点C'处;作∠BPC'的平分线交AB于点E.设BP=x,BE=y,则下列图象中,能表示y与x的函数关系的大致图象是 ( )图X4-6图X4-710.如图X4-8,已知在平面直角坐标系中,直线l1⊥x轴于点A(2,0),点B是直线l1上的动点,直线l2:y=x+1交l1于点C,过点B作直线l3垂直于l2,垂足为D,过点O,B的直线l4交l2于点E.设直线l1,l2,l3围成的三角形的面积为S1,直线l2,l3,l4围成的三角形的面积为S2,且S2=S1,则∠BOA的度数为 ( )图X4-8A.15°B.30°C.15°或30°D.15°或75°二、填空题(每小题4分,共24分)11.分解因式:a2-4b2= .12.二次根式中,x的取值范围是.13.如图X4-9,把正三角形ABC的外接圆对折,使点A落在的中点F处,若BC=6,则折痕在△ABC内的部分DE的长为.图X4-914.如图X4-10,在边长为2的菱形ABCD中,∠ABC=120°,E,F分别为AD,CD上的动点,且AE+CF=2,则线段EF长的最小值是.图X4-1015.如图X4-11,一段抛物线:y=-x(x-3)(0≤x≤3),记为C1,它与x轴交于点O,A1;将C1绕点A1旋转180°得C2,交x轴于点A2;将C2绕点A2旋转180°得C3,交x轴于点A3;…;若P(m,2)在第3段抛物线C3上,则m= .图X4-1116.对于两个不相等的实数a,b,我们规定符号max{a,b}表示a,b中较大的数,如max{2,4}=4.按照这个规定,方程max{x,-x}=的解为.|加加练|1.计算:(-)2+|-4|×2-1-(-1)0.2.解不等式:3x-1≥2(x-1),并把它的解集在数轴上表示出来.图X4-123.化简:+.参考答案1.C2.C3.D4.A5.C6.A7.B8.B9.D10.D11.(a+2b)(a-2b)12.x≤13.414.15.7或816.x=1+或x=-1加加练1.解:原式=3+4×-1=3+2-1=4.2.解:去括号,得3x-1≥2x-2.移项、合并同类项,得x≥-1.把不等式的解集在数轴上表示出来,如图:3.原式=+=+=.选择填空限时练(五)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.下列四个实数,2,0,-1,其中负数是( )A. B.2 C.0 D.-12.下列计算,结果等于a4的是( )A.a+3aB.a5-aC.(a2)2D.a8÷a23.如图X5-1所示,该圆柱体的左视图是( )图X5-1图X5-24.如图X5-3,△ABC内接于☉O,∠A=68°,则∠OBC等于( )图X5-3A.22°B.26°C.32°D.34°5.某校数学兴趣小组在一次数学课外活动中,随机抽查了该校10名同学参加今年初中学业水平考试的体育成绩,统计结果如下表:成绩/分36 37 38 39 40 人数/人 1 2 1 4 2 表中表示成绩的数据中,中位数是( )A.38分B.38.5分C.39分D.39.5分6.用配方法解一元二次方程x2-6x-10=0,变形正确的是 ( )A.(x-3)2=19B.(x+3)2=19C.(x-3)2=1D.(x+3)2=17.不等式组的解集是( )A.x≥2B.1<x<2C.1<x≤2D.x≤28.已知点(-1,y1),(1,0),(3,y2)都在一次函数y=kx-2的图象上,则y1,y2,0的大小关系是( )A.0<y1<y2B.y1<0<y2C.y1<y2<0D.y2<0<y19.如图X5-4,AB是半圆O的直径,半径OC⊥AB于点O,点D是的中点,连结CD,AD,OD,给出以下四个结论:①∠DOB=∠ADC;②CE=OE;③△ODE∽△ADO;④2CD2=CE·AB.其中正确结论的序号是( )图X5-4A.①③B.②④C.①④D.①②③10.如图X5-5,矩形ABCD中,AB=8,BC=6,将矩形ABCD绕点A按逆时针方向旋转得到矩形AEFG,AE,FG分别交射线CD于点P,H,连结AH.若P是CH的中点,则△APH的周长为 ( )图X5-5A.15B.18C.20D.24二、填空题(每小题4分,共24分)11.分解因式:a2-4a= .12.一个布袋里装有10个只有颜色不同的球,其中红球有m个,从布袋中随机摸出一个球记下颜色后放回、搅匀,再摸出一个球,通过大量重复试验后发现,摸到红球的频率稳定在0.3左右,则m的值为.13.某种品牌手机经过4,5月份连续两次降价,每部售价由5000元降到3600元.已知5月份降低的百分率是4月份降低的百分率的2倍,设4月份降低的百分率为x,根据题意可列方程: .14.如图X5-6,用一个半径为60 cm,圆心角为150°的扇形围成一个圆锥,则这个圆锥的底面半径为cm.图X5-615.如图X5-7,把菱形ABCD沿折痕AH翻折,使B点落在BC延长线上的点E处,连结DE.若∠B=30°,则∠CDE= °.图X5-716.如图X5-8,直角坐标系xOy中,直线y=-x+b分别交x,y轴的正半轴于点A,B,交反比例函数y=-的图象于点C,D(点C在第二象限内),过点C作CE⊥x轴于点E,记四边形OBCE的面积为S1,△OBD的面积为S2,若=,则CD的长为.图X5-8|加加练|1.计算:(-2)0-()2+|-1|.2.解不等式组:3.解方程:-1=.参考答案1.D2.C3.C4.A5.C6.A7.C8.B9.C10.C11.a(a-4)12.313.5000(1-x)(1-2x)=360014.2515.4516.5加加练1.解:原式=1-6+1=-4.2.解:解不等式①,得x>-3,解不等式②,得x<5,∴不等式组的解是-3<x<5.3.解:原方程可化为2-(x-2)=3x,解得x=1.经检验,x=1是原方程的解.所以原方程的解是x=1.选择填空限时练(六)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.下列四个数中,是正整数的是 ( )A.-1B.0C. D.12.据媒体报道,我国因环境污染造成的巨大经济损失,每年高达680000000元,这个数用科学记数法表示正确的是( )A.6.8×109元B.6.8×108元C.6.8×107元D.68×107元3.下列事件中,必然事件是 ( )A.今年夏季的雨量一定多B.下雨天每个人都打着伞C.二月份有30天D.我国冬季的平均气温比夏季的平均气温低4.如图X6-1,点A,B,C,D,O都在方格纸的格点上,若△COD是由△AOB绕点O按逆时针方向旋转而得,则旋转的角度为( )图X6-1A.30°B.45°C.90°D.135°5.一次函数y=2x-2的图象不经过的象限是( )A.第一象限B.第二象限C.第三象限D.第四象限6.下列四个图形,其中是轴对称图形,且对称轴的条数为2的图形的个数是( )图X6-2A.1个B.2个C.3个D.4个7.对于反比例函数y=,下列说法不正确的是( )A.点(-3,-1)在它的图象上B.它的图象在第一,三象限C.y随x的增大而减小D.当x>1时,y<38.如图X6-3,在菱形ABCD中,对角线AC,BD相交于点O,E为BC的中点,则下列式子一定成立的是( )图X6-3A.AC=2OEB.BC=2OEC.AD=OED.OB=OE9.如图X6-4,将长为2,宽为1的矩形纸片分割成n个三角形后,拼成面积为2的正方形,则n≠( )图X6-4A.2B.3C.4D.510.小阳在如图X6-5①的扇形舞台上沿O➝M➝N匀速行走,他从点O出发,沿箭头所示的方向经过点M再走到点N,共用时70秒.有一台摄像机选择了一个固定的位置记录了小阳的走路过程,设小阳走路的时间为t(单位:秒),他与摄像机的距离为y(单位:米),表示y与t的函数关系的图象大致如图X6-5②,则这个固定位置可能是图X6-5①中的( )图X6-5A.点QB.点PC.点MD.点N二、填空题(每小题4分,共24分)11.使代数式有意义的x的取值范围是.12.东山茶厂有甲、乙、丙三台包装机,同时分装质量为200克的茶叶.从它们各自分装的茶叶中分别随机抽取了15盒,测得它们的实际质量的方差如下表:甲包装机乙包装机丙包装机方差(克2) 5.6 9.3 0.9 根据表中数据,三台包装机中, 包装机包装的茶叶质量最稳定.13.如图X6-6,l1是反比例函数y=在第一象限内的图象,且过点A(2,1),l2与l1关于x轴对称,那么图象l2的函数解析式为(x>0).图X6-614.将一个三角形经过放大后得到另一个三角形,如果所得三角形在原三角形的外部,这两个三角形各对应边平行且距离都相等,那么我们把这样的两个三角形叫做“等距三角形”,它们对应边之间的距离叫做“等距”.如果两个等边三角形是“等距三角形”,它们的“等距”是1,那么它们周长的差是.15.已知在平面直角坐标系内,以点P(1,2)为圆心,r为半径画圆,☉P与坐标轴恰好有三个交点,那么r的取值是.16.在平面直角坐标系xOy中,抛物线y=-x2+2mx-m2-m+1交y轴于点A,顶点为D,对称轴与x 轴交于点H.(1)顶点D的坐标为(用含m的代数式表示);(2)当抛物线顶点D在第二象限时,如果∠ADH=∠AHO,那么m的值为.|加加练|1.计算:3-2-2cos60°+(12-2006)0-|-|.2.先化简,再求值:(1-)÷,其中x请从-2,-1,0,1,2中选一个恰当的数.参考答案1.D2.B3.D4.C5.B6.C7.C8.B9.A10.B11.x≥-112.丙13.y=-14.615.或216.(1)(m,1-m)(2)m=-1或m=-2加加练1.解:原式=-2×+1-=-.2.解:原式=÷=·=x+2.∵x≠0,1,-2,∴x可取-1或2.当x=2时,原式=2+2=4.(或当x=-1时,原式=-1+2=1)选择填空限时练(七)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.计算(-6)+5的结果是( )A.-11B.11C.-1D.12.函数y=中,自变量x的取值范围是( )A.x≠2B.x≥2C.x>2D.x≥-23.在以下“绿色食品”“节能减排”“循环回收”“质量安全”四个标志中,是轴对称图形的是( )图X7-14.如图X7-2是由4个相同的正方体搭成的几何体,则其俯视图是( )图X7-2图X7-35.一个不透明的布袋中有2个白球,3个黑球,除颜色外其他都相同,从中随机摸出一个球,恰好为黑球的概率是( )A. B. C. D.6.如图X7-4,矩形ABCD的两条对角线交于点O,若∠AOD=120°,AB=6,则AC等于( )图X7-4A.8B.10C.12D.187.不等式2(x-1)≥x的解在数轴上表示为( )图X7-58.如图X7-6,已知D,E分别是△ABC的边AB,AC上的点,DE∥BC,且BD=3AD,那么AE∶AC等于( )图X7-6A.2∶3B.1∶2C.1∶3D.1∶49.如图X7-7,已知正方形ABCD的边长为1,分别以顶点A,B,C,D为圆心,1为半径画弧,四条弧交于点E,F,G,H,则图中阴影部分的外围周长为( )图X7-7A.πB.πC.πD.π10.把三张大小相同的正方形卡片A,B,C叠放在一个底面为正方形的盒底上,底面未被卡片覆盖的部分用阴影表示.若按图X7-8①,②摆放,阴影部分的面积分别为S1和S2,则S1和S2的大小关系是( )图X7-8A.S1=S2B.S1<S2C.S1>S2D.无法确定二、填空题(每小题4分,共24分)11.分解因式:ab-2a= .12.已知一组数据:2,1,-1,0,3,则这组数据的中位数是.13.在同一平面直角坐标系内,将函数y=2x2-3的图象向右平移2个单位,再向下平移1个单位后得到新图象的顶点坐标是.14.如图X7-9,将Rt△ABC绕直角顶点A顺时针旋转90°,得到△AB'C',连结BB',若∠1=25°,则∠C的度数是.图X7-915.如图X7-10,在平面直角坐标系中,直线y=kx+b与x轴,y轴分别交于点A(4,0),B(0,2),点C为线段AB上任意一点,过点C作CD⊥OA于点D,延长DC至点E使CE=DC,作EF⊥y轴于点F,则四边形ODEF的周长为.图X7-1016.如图X7-11,已知AB,CD是☉O的两条相互垂直的直径,E为半径OB上一点,且BE=3OE,延长CE交☉O于点F,线段AF与DO交于点M,则的值是.图X7-11|加加练|1.计算:-2cos 45°+()-1.2.化简:+.3.求满足不等式组的所有整数解.参考答案1.C2.B3.A4.A5.C6.C7.C8.D9.B10.A11.a(b-2)12.113.(2,-4)14.70°15.816.加加练1.解:原式=2-2×+2=+2.2.解:原式====2.3.解:解x-3(x-2)≤8,得x≥-1,解x-1<3-x,得x<2,所以不等式组的解集为-1≤x<2,其中所有的整数解为-1,0,1.选择填空限时练(九)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.在-2,0,,1这四个数中,最大的数是( )A.-2B.0C.D.12.如图X9-1是由几个相同的小正方体搭成的一个几何体,它的俯视图是( )图X9-1图X9-23.抽样调查某公司员工的年收入数据(单位:万元),结果如下表:年收入/万元 5 6 7 15 30 人数8 6 3 2 1 则可以估计该公司员工中等年收入约为( )A.5万元B.6万元C.6.85万元D.7.85万元4.C919大型客机是中国具有自主知识产权的干线民用飞机,其零部件总数超过100万个,将100万用科学记数法表示为( )A.1×106B.100×104C.1×107D.0.1×1085.如图X9-3,AB是☉O的弦,OC⊥AB,交☉O于点C,连结OA,OB,BC,若∠ABC=20°,则∠AOB的度数是( )图X9-3A.40°B.50°C.70°D.80°6.不等式的解x≤2在数轴上表示为 ( )图X9-47.如图X9-5,在△ABC中,两条中线BE,CD相交于点O,则S△DOE∶S△COB等于( )图X9-5A.1∶2B.1∶3C.1∶4D.2∶38.小明进行两次定点投篮练习,第一次a投b中(a≥b),第二次c投d中(c≥d),用新运算“ ”描述小明两次定点投篮总体命中率,则下列算式合理的是( )A. =B. =C. =D. =9.如图X9-6,抛物线y1=-(x+2)2-1与y2=a(x-4)2+3交于第四象限点A(1,-4),过点A作x轴的平行线,分别交两条抛物线于B,C两点,且D,E分别为顶点.则下列结论正确的是 ( )图X9-6A.AB<ACB.当x>1时,y1>y2C.△ACE是等边三角形D.△ABD是等腰三角形10.如图X9-7,菱形ABCD中,∠ABC=60°,边长为3,P是对角线BD上的一个动点,则BP+PC的最小值是( )图X9-7A. B.C.3D.+二、填空题(每小题4分,共24分)11.分解因式:2m2-8= .12.如图X9-8,把一张长方形纸带沿着直线GF折叠,若∠CGF=30°,则∠1的度数是.图X9-813.某城市为了了解本市男女青少年平均身高发育情况,随机调查了6岁~18岁男女青少年各100人,制作成如图X9-9所示的不同年龄平均身高统计图,从图中可知,该城市的男性青少年的身高高于同年龄女性的年龄段大概是.图X9-914.如图X9-10,P是边长为a的等边三角形ABC内任意一点,过点P分别作三角形三边的垂线PD,PE,PF,垂足分别点为D,E,F,则图中阴影部分图形的面积总和为(用含a的式子表示) .图X9-1015.如图X9-11,正方形ABCD的边长为4,在这个正方形内作等边三角形EFG,使它们的中心重合,则△EFG的顶点到正方形ABCD的顶点的最短距离是.图X9-1116.下面是一种算法:输入任意一个数x,都是“先乘2,再减去3”,进行第1次这样的运算,结果为y1,再对y1实施同样的运算,称为第2次运算,结果为y2,这样持续进行,要使第n次运算结果为0,即y n=0,则最初输入的数应该是.(用含有n的代数式表示)|加加练|1.化简:÷(-1).2.[2018·成都 ]为了美化环境,建设宜居成都,我市准备在一个广场上种植甲、乙两种花卉.经市场调查,甲种花卉的种植费用y(元)与种植面积x(m2)之间的函数关系如图X9-12,乙种花卉的种植费用为每平方米100元.(1)直接写出当0≤x≤300和x>300时,y与x的函数关系式.(2)广场上甲、乙两种花卉种植面积共1200 m2,若甲种花卉的种植面积不少于200 m2,且不超过乙种花卉种植面积的2倍,那么应该怎样分配甲、乙两种花卉的种植面积才能使种植总费用最少?最少费用为多少元?1.D2.D3.B4.A5.D6.B7.C8.C9.D10.B[解析] 如图,过点P作PM⊥AB于点M,过点C作CH⊥AB于点H.∵四边形ABCD是菱形,∠ABC=60°,∴∠PBM=∠ABC=30°,∴PM=PB,∴PB+PC=PC+PM.根据垂线段最短可知,CP+PM的最小值为CH的长.在Rt△CBH中,CH=BC·sin 60°=,∴PB+PC的最小值为.故选B.11.2(m+2)(m-2)12.60°13.6~10岁和14~18岁14.15.4-216.加加练1.解:原式=÷=·=.2.解:(1)当0≤x≤300时,设函数关系式为y=k1x,过(300,39000),则39000=300k1,解得k1=130.∴当0≤x≤300时,y=130x;当x>300时,设函数关系式为y=k2x+b,过(300,39000)和(500,55000)两点,∴解得∴y=80x+15000.综上y=(2)设甲种花卉的种植面积为a m2,则乙种花卉的种植面积为(1200-a) m2.根据题意得解得200≤a≤800.当200≤a≤300时,总费用W1=130a+100(1200-a)=30a+120000,当a=200时,总费用最少为W min=30×200+120000=126000(元);当300<a≤800时,总费用W2=80a+15000+100(1200-a)=-20a+135000,当a=800时,总费用最少为W min=-20×800+135000=119000(元).∵119000<126000,∴当a=800时,总费用最少,为119000元,此时1200-a=400.答:当甲、乙两种花卉种植面积分别为800 m2和400 m2时,种植总费用最少,最少费用为119000元.。

2021年天津中考英语综合填空专练3(共5道打印版)

2021年天津中考英语综合填空专练3(共5道打印版)

2021年天津中考英语综合填空专练3（共5道打印版）八、综合填空（本大题共10小题，每小题1分，共10分）根据短文内容及首字母提示，填写所缺单词，使短文意思完整。

As teenagers, you have many dreams. These dreams can be very b________(81), such as winning the Nobel Prize, or they can be small, you may just want to be the b________(82) student in your class. Once you find a dream, what do you do with it? Do you ever try to m________(83) your dream real?Andrew Matthews, an Australian writer tells us that making our dreams real is life’s biggest challenge(挑战). You may think y ou’re not very good at some school s_______ (84)or it is impossible for you to become a writer. These kinds of thoughts s________(85) you from getting your dream. In fact, everyone can make his dream come true. The first thing you must do is to r________(86) what your dream is. Don’t let it leave your heart. Keep telling y_________(87)what you want. Do this step by step and your dream will come true faster b________(88) a big dream is made up of many small dreams.You must also n_________(89) give up your dream. There will be difficulties on the road to your dreams. But the biggest difficulty comes from yourself. You need to decide w___________(90) is the most important.Without dreams, you won’t make up your mind to learn more skills and find new interests.81. __________ 82. __________ 83. __________84. __________85. __________86. __________ 87. __________ 88. __________89. __________90. __________八、综合填空（共10分，每小题1分）阅读下面的短文，根据每个空格中所给的词首字母填入适当的词，使短文意思完整。

2021年北京市中考数学总复习选择填空提分特训(03)

2021年北京市中考数学总复习选择填空提分特训(03)

选择填空提分特训(三)[限时:25分钟满分:32分]一、选择题(每小题2分,共16分)1.下列立体图形中,主视图是圆的是()图X3-12.将一副直角三角板如图X3-2放置,那么∠AOB的大小为()图X3-2A.150°B.135°C.120°D.90°3.电影《流浪地球》中,人类计划带着地球一起飞向约4光年外的新家园,已知光年是天文学中的距离单位,1光年大约是95000亿千米,则4光年约为()A.9.5×104亿千米B.95×104亿千米C.3.8×105亿千米D.3.8×104亿千米4.甲、乙、丙、丁四名工人一天中生产零件的情况如图X3-3所示,每个点的横、纵坐标分别表示该工人一天中生产I型、Ⅱ型零件数,则四名工人中日生产零件总数最大的是()图X3-3A.甲B.乙C.丙D.丁5.如果a-b=√3,那么代数式b2a -a·aa+b的值为()A.-√3B.√3C.3D.2√36.实数a,b,c,d在数轴上的对应点的位置如图X3-4所示,则下列结论正确的是()图X3-4A.|a|>|b|B.a>-3C.a>-dD.1c<17.在图X3-5所示的图形中,根据尺规作图的痕迹,能判断射线AD平分∠BAC的是()图X3-5A.图②B.图①与图②C.图①与图③D.图②与图③8.某市为了解旅游人数的变化情况,收集并整理了2017年1月至2019年12月期间的月接待旅游量(单位:万人次)的数据并绘制了统计图如图X3-6,根据统计图提供的信息,下列推断不合理的是()图X3-6A.2017年至2019年,年接待旅游量逐年增加B.2017年至2019年,各年的月接待旅游量高峰期大致在7,8月份C.2019年的月接待旅游量的平均值超过300万人次D.2017年至2019年,各年下半年(7月至12月)的月接待旅游量相对于上半年(1月至6月)波动性更小,变化比较平稳二、填空题(每小题2分,共16分)9.若分式x的值为正,则实数x的取值范围是.x2+210.估计无理数√11在连续整数与之间.11.在平面直角坐标系中,把点A(2,3)向左平移一个单位得到点A',则点A'的坐标为.12.如图X3-7,☉O的半径为4,△ABC是☉O的内接三角形,连接OB,OC.若∠BAC与∠BOC互补,则弦BC的长为.图X3-713.《九章算术》是我国古代数学的经典著作,书中有一个问题:“今有黄金九枚,白银一十一枚,称之重适等,交易其一,金轻十三两,问金、银一枚各重几何?”意思是:甲袋中装有黄金9枚(每枚黄金质量相同),乙袋中装有白银11枚(每枚白银质量相同),称重两袋相等,两袋互相交换1枚后,甲袋比乙袋轻了13两(袋子的质量忽略不计),问黄金、白银每枚各重多少两?设每枚黄金重x两,每枚白银重y两,根据题意可列方程组为.14.如图X3-8所示的正方形网格中,∠1∠2(填“>”“<”或“=”).图X3-815.菱形、矩形与正方形的形状有差异,我们将菱形、矩形与正方形的接近程度称为菱形或矩形的“接近度”.设菱形相邻两个内角的度数分别为m°,n°.若我们将菱形的“接近度”定义为mn(m<n),则菱形的“接近度”=时,菱形就是正方形.图X3-916.在菱形ABCD中,M,N,P,Q分别为边AB,BC,CD,DA上的点(不与端点重合).对于任意菱形ABCD,下面四个结论中,①至少存在一个四边形MNPQ是正方形;②存在无数个四边形MNPQ是矩形;③存在无数个四边形MNPQ是菱形;④存在无数个四边形MNPQ是平行四边形.所有正确结论的序号是.附加训练17.解不等式组:{2(x-1)<x, x-53<x+1.18.如图X3-10,△ABC中,AB=BC,D在BC的延长线上,连接AD,E为AD的中点.(1)尺规作图:作∠ABC的平分线,与线段AC交于点F,连接EF;(2)根据(1)中所作的图形,证明:EF∥BC.图X3-10【参考答案】1.D2.B3.C4.C[解析]四名工人中,丙的横、纵坐标的和最大,即日生产零件总数最大,故选C.5.A6.A7.C8.D[解析]从折线统计图的整体变化情况可得2017年至2019年,年接待旅游量逐年增加,因此选项A不符合题意;2017年至2019年,各年的月接待旅游量高峰期大致在7,8月份,因此选项B不符合题意;从2019年3月起,每个月的旅游量均超过300万人次,并且整体超出的还很多,因此选项C不符合题意;从统计图中可以看出2017年至2019年,各年下半年(7月至12月)的月接待旅游量相对于上半年(1月至6月)波动性要大,因此选项D符合题意.故选D.9.x>010.3411.(1,3)12.4 √313.{9x=11y,(10y+x)-(8x+y)=1314.>15.1 16.①②③④17.解:{2(x-1)<x,①x-53<x+1.②解不等式①,得:x<2,解不等式②,得:x>-4.∴原不等式组的解集为-4<x<2.18.解:(1)如图所示:(2)证明:∵AB=BC,BF平分∠ABC,∴F是AC的中点.∵E是AD的中点,∴EF是△ACD的中位线.∴EF∥BC.。

2021年英语中考选词填空专项训练64篇(含答案)

2021年英语中考选词填空专项训练64篇(含答案)

2021年英语中考选词填空专项训练64篇（含答案）(1)city, move, good, either, or, friend, with, than, improve, quietIn China many people are leaving the countryside to work in the cities．They know that the countryside is much poorer1．the city．And there isn't much work there．Services such as hospitals and schools are much2．in the city than in the countryside. They hope that their lives will3．when they move to the city. But in many big4．like London or Paris，people are moving out. They want to live a 5．life in the countryside．They don't like the noise of the city. They don't like the crowded trains and buses, 6．．They don't want to live in the cities any more．They want a house7．a garden in the countryside because there is fresh air.For the people who move from the city to the countryside，not all of them are happy．Two8．three years later，many people find that they don't make much money and there isn't much work to do. People in the countryside are very different and they aren't always very 9．. As a result, a lot of people who have 10．to the countryside move back to the city again．(2)among, to, study, test, correct, between, low ,country, be, much, by, alsoDo you know how well your country speaks English? China's English proficiency ranks(排名) 40th, according 11．a report from Swedish education company Education First (EF).China rose seven places from last year.The company 12．more than 2.3 million adults in 100 non-English-speaking countries and regions(地区). China scored 53.44 out of 100 points. For the first time, Chinese people's English has reached a moderate level(中等水平) compared to other non-English-speaking 13．.People with this English level can read foreign literary(文学的) works and science articles in English. They can 14．express their views and discuss academic issues (学术问题) with others in English 15．, according to China’s Stand ards of English Language Ability, which was released by the Ministry of Education last year.16．all non-English-peaking countries, adults from the Netherlands were the best English speakers with a score of 70.27. The country with 17．score is Libya at 40.87.There 18．a few reasons for China’s rise in the rankings, says Jesper Knutella, executive vice-president of EF. Chinese people have a strong desire (渴望) to see and understand the world, which pushes them to study English.Also, they spend large amounts of time and money 19．English in the long term. Last year, 300 million people in China were learning English, according to the Chinese Academy of Social Sciences. The market was worth as 20．as 500 billion yuan.(3)she worry about never work many young much see feel give oldMother love is true love. It gives everybody everything all his life. When you are still a baby, Mother takes good care of you as 21．as possible. In your waking hours, she always holds you in22．arms. When you are ill, she stops her23．to look after you day and night and forgets about herself. When you grow up day by day, she24．very happy. When you are25．enough to go to school, Mother still looks after you all the time. On cold winter days, she always tells you to put on26．clothes. She always stands in the wind waiting for you back from school. When you hurry to leave home for school with little breakfast, she is always27．about you at home. She usually knows28．your study and spends much money on your school things. When you do well at school, the brightest smile will be29．on her face.Mother is always ready to30．everything that she has to her children, not to receive. What true love that is in the world! We will remember Mother Love forever.(4)thank realize excite look several lose gift disappear everything but Louis was a child. He always stayed at home playing computer games.One day, he went hiking with his classmates. He got31．in a forest and saw a dwarf（侏儒）. The dwarf didn’t notice a huge stone was rolling towards him. Louis saved the dwarf. The dwarf32．Louis so much that he took Louis to a well. “This is a magic well,” said the dwarf. “Wonderful33．often come out of it, and youcan enjoy them.”They waited for a moment and a cool bike came out. Louis was very 34．. He got on it to have a try. However, the bike35．soon. Then, a beautiful jacket came out. Again, Louis spent a long time36．at it. Just after he put it on, the jacket disappeared. This happened37．times. At first Louis was angry, but he soon tried his best to enjoy these gifts.When he sat down and had a rest, the dwarf explained that the well had always worked this way. It was like life. There are many good things in life, 38．they didn’t last for a long time.Hearing his words, Louis39．his problem. He spent all his time playing games and didn't enjoy the good things in life. Since then, Louis learned to see 40．as a gift and tried to enjoy it. He didn't feel bored anymore.(5)So surprised I shout quick pull on friend feel rockI don't like heights. I mean I don't mind being in a tall building or on a plane. However, looking out over the edge of a cliff makes me 41．nervous.A few years ago,I went on a boat trip with some of my 42．. On the way, we came across a cliff with a jumpi ng rock. One of my friends shouted, “Hey, the 43．height is perfect for jumping.” I felt nervous because I had never told them about my fear of heights.Everybody jumped off the rock many times except 44．. “Hey John! You haven't jumped yet. Are you scared?” joked one of the boys. Soon, my friends were all 45．, “Jump, jump, jump!”I was too embarrassed to say I was afraid, 46．I climbed the rock. Everyone was watching as I looked down. Only Peter was in the water far below and I just couldn't do it. Suddenly, a rock fell and hit Peter 47．the head. I called him, but he didn't answer. Without thinking twice,I jumped into the water and swam to him 48．. Then, Peter was 49．by me to the shore. He coughed a little, but he was fine.I still don't like heights, but I'm still 50．at what I did that day. I guess we never know what we' re capable of when someone needs our help.Word: cliff峭壁；embarrassed尴尬的；shore岸；capable能胜任的(6 )however for tradition to there few invent nearly television one discovers big Television plays an important part in our lives. Throughout the world, it entertains (使快乐) and informs (告发) people of all ages. It is also used 51．advertising (做广告). But how did it begin?The first black and white television 52．in the 1920s. In the early days, only a few people could afford televisions. In 1936, 53．were only about 200 TVs in the world. Later, the price came down and more people bought one. Then, in the USA in the 1950s, color televisions appeared, but 54．people bought them at first because they could not show black and white programs. 55．, a new kind of television, was invented in 1967. This could show black and white and color programs, so people started buying these new TVs.Advertising on TV started in 1941 in the USA．The 56．advertisement was for a kind of watch. Today advertisements are seen on 57．all channels (频道) all over the world.In China, the first television was made in 1958. It was a black and white television and 8ppeared in Tianjin. Now 58．are in nearly every home in China.With new technology, television is facing a 59．challenge. Today more and more people choose to watch TV programs online. They prefer to use their phones or computers instead of televisions.If this manner of watching shows keeps rising, will the 60．channels and television be forgotten? We can only wait and see what the future has in store for us.(7)she and hurt never but sad fly good angry anything change money Long long ago, there was a swan(天鹅) with golden feathers(金羽毛), She lived in a lake.A woman lived in a small, house near the lake with her two daughters. They were very poor. They worked hard all year round, 61．still, they lived a hard life and sometimes they even didn't have enough 62．to buy food.The swan was 63．to see that. She said to herself, “I’ll give one of my feathers to them each day, and then they can live a happy life with the money selling my feathers. “That evening, she 64．to the poor woman’s house and left a golden feather on the table without saying 65．.From then on, the swan came every day and gave them a feather. The woman was very happy because their life was much 66．than before.But day after day，the woman became greedy (贪婪的).She said to her daughters, “The swan may flyaway one day. If so, we will be poor again. We should take all 67．feathers when she comes next time.“Oh, no, Mom!” cried the daughters. “This will 68．the swan. She helps us a lot!” “But the mother wouldn’t listen. When the swan came as usual, the mother caught her and took her feathers. But suddenly, the golden feathers 69．into chicken feathers.Then, the golden swan said, “Poor Mother, I came to help you, but you wanted to kill me. Now I’m leaving and will 70．come back. Don’t be greedy!”(8)strict difficulty easy allow cheer depend decide my upset ownHave you ever lived in a school dormitory? Some students think living at school is interesting. Well, it's not as71．as they might imagine.Last year, my parents decided to let me live at school. I was surprised to hear this72．. I could be in control of myself without my parents managing me.However, when I first arrived, I got anxious about the73．rules. We had to study from 7 a.m. to 7 p.m. every day. We were not74．to play with cell phones or go back to the dormitory to take showers during this time. I also argued with one of my roommates. These things made me feel75．and homesick. I just wanted to go back home.On weekends, I went home. My mother talked to me, "Don't be afraid. Try to think about how to improve the situation." My mother's words76．me up. None of my roommates had such problems. I should solvemy77．problems.The next week, I tried my best to calm78．down and focus on my work. It did not seem too hard. Then I talked with my roommates in a friendly way and make more friends. I realized everything79．on our own attitudes. I begin to enjoy the school life.Every time I fight with80．, I always think of this special experience. It gives me courage to overcome them. Perhaps I have grown up through this.(9)when umbrella they interesting where taste or much leaf late boring know There are 24 solar terms (节气) in China, and the Grain Rain (谷雨) is one of 81．. It falls on April 19 this year.This solar term is called the Grain Rain because it is 82．for “rain” that helps the grain grow. From its name, we can guess it means a time of 83．rain. So if you go out in the next two weeks，you may need to bring an 84．with you.But this is a great time for planting crops. If you miss the time for planting during the Grain Rain, you will feel sorry 85．.So maybe you can try to plant some flowers 86．vegetables in your garden.87．the spring rain falls, farmers begin to grow crops. Farmers often say, “Spring rain is as precious (珍贵) as oil.” It brings farmers hope for a good year ahead.There are also many 88．customs during the Grain Rain. In the northern part of China, people like to eat Chinese toon(椿芽) mixed with eggs. The food made during the Grain Rain 89．wonderful and is good for your stomach. In the southern part of China, the tea 90．are picked during the time. They are called Grain Rain tea, which is famous for its freshness and sweet smell.(10)trouble in weak easy turn usual you meet but howEven though your needs are greater when you have big trouble, it can be hard to ask for help to 91．those needs.To get the help you need, think about 92．to family and friends, others who also have 93．, people you meet in support groups and even help providers.No one needs to face big trouble alone. When people in trouble look for and receive help from others, they often find it 94．to live through.You may find it hard to ask for or accept help. After all, you are used to taking care of yourself. Maybe you think that asking for help shows 95．. Or perhaps you do not want to let others know that some things are hard for you to do. All these feelings are not 96．. As one man with a great difficulty once said, "I had always been the strong one. Now I had to turn to others for help. It wasn't easy at first, 97．the support of others helped me get through a lot of hard times."People feel good when they help others. Your friends may not know what to say or 98．to act when they are with you. Some people may even avoid you. But they may feel better when you ask them to cook a meal orpick up 99．child after school. There are many ways that family, friends, other people who have similar troubles and many others can help. 100．turn, there are also ways you can help and support them!(11)cheap frighten arrive I passenger everywhere rider truck follow take Have you ever been to London? Maybe you haven’t.Now, I’ll tell you something about it. If you go to visit London, you’ll see not only a lot of buses and cars, but also lots of bikes on the road. Why are there so many bikes?First, it’s very 101．to buy a bike. Second, riding a bike can save much time. You often have to wait for a bus for half an hour. What’s more, there are so many 102．waiting for the bus. Even though you get on the bus, it’s not easy to find a seat to sit down.I went to work by bus for about four years. I often 103．at my office late and tired. Then one day, about two years ago, a friend of 104．said, “I go to work by bike. Why don’t we go together?”“My bike is old,” I answered, “and there are so many cars and 105．on the roads. I’ll feel 106．.”“You needn’t feel frightened,” said my friend. “If you 107．me and we ride slowly, you’ll be fine.”The next day I bought myself a new bike. Although we went slowly, we arrived at work early. It 108．me 40 minutes to go by bus, but only half an hour by bike!Now I don’t feel afraid. I love going to work by bike.Many 109．think in the same way as I do. That’s why you see bikes 110．in the city. Next year you’ll see more, I am sure. Perhaps in the future we will have roads for bikes only. I hope so. Then the world will bea world of bikes.(12)good be but two astronaut study information spend or sheEllen Ochoa is the first Hispanic(西班牙裔)astronaut woman to fly into space. As a child, Ochoa didn't know what she wanted to be when she grew up. 111．her mother always told her stay in school and study hard. She listened to her mother. When she was 13 years old, she won a spelling contest. In high school, Ochoa spent many hours each week 112．for all her subjects. She finished with the 113．grades in her class. After high school, Ochoa went to college. One day in college she heard that the U. S. government had chosen six women to become 114．. After her many years of hard work. She115．asked to be an astronaut in 1990. In April, 1993 in the space shuttle（航天飞机）named Discovery, she made history by becoming the first Hispanic woman to travel to space. Ochoa and the other astronauts had many jobs while on the Discovery. One of116．tasks was to use the shuttle's robot arm to love large objects. Ochoa used a computer to make the arm pick up a satellite and put it into space. A satellite is a spacecraft that moves around the earth 117．some other objects. Satellites can take pictures and gather(收集）118．about planets and stars. During her 119．trip in 1994, she used the robot arm again. She used the arm to pull in a satellite from space. The satellite had peen used to study the air around the earth. Then Ochoa went to two more shuttle trips. Altogether, she 120．a whole month in space. That's more than700 hours. She is a hero to many young people. She often tells young people: Nothing is impossible if you put your heart into it.(13)surprise but regret friend continue without early depend everything they Do you know why children can't look after themselves these days? The main reason is that their parents help them do 121．. This has caused teenagers to 122．too much on their parents. Therefore, some teenagers may do well in school, but they still lack independence (缺乏独立性).Here is a story. Last year, a seventeen-year-old boy got into a good university. But one month later, he dropped out of school. His parents were 123．and asked him why he had come back home. He answered, “I realized that I lack independence. I can't live alone 124．you. I hate this feeling. My classmates and 125．laugh at me. They think I am like a little boy and they say I need my mom to help me do everything.” Then, his parents realized their son wasn't kidding. They 126．how they had raised him. Teenagers need to be independent. Otherwise(否则), they will have trouble 127．their education and even living on their own.Parents should love their children, 128．spoiling(溺爱) them is bad for them. Some grandparents spoil their grandchildren very much. They think their grandchildren should get everything they want. But they don't know how important it is for 129．grandchildren to be independent. The 130．kids learn to be independent, the better off they will be in life. Parents should let go in the right time.(14)usual, common, spread, However, type, as, cause, touch, specific, through The new coronavirus(新型冠状病毒) that is 131．in the world is in the same family of infectious (传染的) diseases 132．SARS (非典) and MERS (中东呼吸综合症).Coronaviruses may 133．diseases in animals and humans, but animals are 134．the origin (来源) of coronaviruses. There are different 135．of coronaviruses that can cause different symptoms. Some symptoms are mild, just like a 136．cold. Patients may have a runny nose, cough, sore throat and fever. 137．, some types of coronaviruses can cause severe diseases, such as breathing problems and pneumonia (肺炎).Coronaviruses can spread from person to person. People may catch the viruses 138．the coughing and sneezing of an infected (被感染的)person or by touching things that an infected person 139． . There is no vaccine (疫苗) or 140．treatment for coronaviruses right now.(15)wealth, how, money, opposite, marry, refuse, in, common, offer, so There was once a farmer. He had a very beautiful daughter. The girl was 141．beautiful that she attracted many young men in her neighborhood. One day, her father asked her, “My dear daughter, tell me, what kin d of man do you want to be your husband?” The girl said, “My future husband can be poor but also a rich man.” “142．could that be? Poor and rich are two 143．things!” said her father. “Dad, a poor person can also have his 144．.” said his daughter. The fath er then told people that his daughter was ready to 145．.One by one, the men came forward but the girl 146．them all. The father began to worry. Suddenly, a young man who dresses very 147．appeared. The father asked, “Young man, what can you 148．my daught er?” The young man said, “My wealth (财富) is always with me. It is my hands. I am a good carpenter (木匠). I can make tables and chairs 149．an hour. I can cook, too. However, I do not have much 150．. But, with my pair of hands, I do have a whole life of wealth ahead of me!” “That’s great!” shouted the girl excitedly. “You are my idea of a dream husband!”A pair of hardworking hands can create much wealth.(16)make, here, fun, do, healthy, lock,attention, forget, welcome, they, support, brave, Wuhan is the capital city of Hubei. It is a beautiful, large and very important city in China. During this Spring Festival, the city received tons of 151．because a novel coronavirus(新型冠状病毒) appeared there and spread around the country.Several days before the Spring Festival, Wuhan was 152．and the city became quiet and still. However, a big fight was going on there It was the fight against the novel coronavirus. The doctors and nurses there became 153．warriors(勇士) in masks, glasses and protective suits. They were racing against the clock. Lots of 154．had to work over ten hours every day, drinking little or no water. They risked their 155．and even lives to treat the patients. In such cases, almost no one stepped back or gave up. Thousands of doctors and nurses from other cities and provinces in China went to Wuhan 156．them. There is no doubt that they were all 157．in the fight.Besides doctors and nurses, everyone in our country was 158．their own part in the fight against the novel coronavirus.The Spring Festival was not as 159．as it usually is, It even brought us to tears many times. However, we will never 160．those difficult days, not only for the awful virus, but also for the efforts we made together and the lessons we learned.(17)help, kind, on, Canada, they, give, always, invent, so, resultMr King owns a restaurant in a big city. He is a 161．. And people are always thinking he is very professional in trading. The prices of the dishes are 162．low that many people enjoy having dinner in his restaurant. So the restaurant is 163．very busy.164．the other hand, a modern robot has played an important role in the restaurant. The robot is used for cooking and it is made of metal and wood. It 165．by some college students. It can make some cookies, pies and many 166．of dishes. And the taste of the dishes is very delicious. As a 167．, Mr. King has made a lotof money because of the good business.Mr. King has already 168．away some of his money to lots of homeless people. When he is mentioned by the local people, he is a great hero in 169．hearts. Mr. King said he used to be very poor when he was young, so he made a decision 170．the poor when he could make a lot of money. It is always his dream, and now his dream has come true.(18)what, leave, achieve, become, to, change, possible, heart, which, comeAs teenagers, you have many dreams. These dreams can be very big, or they can be small. You may just want to 171．one of the top ten students in your class.You may think that you're not very good at some school subjects, or that it is 172．for you to become a writer. These kinds of thoughts prevent you from 173．your dream.In fact, everyone can make their dream come true. The first thing you must do is to remember the dream in your mind. Don't let it 174．your heart. Keep telling yourself 175．you want. Do this step by step and your dream will come true more quickly, because a big dream is made up of many small ones. There will be difficulties on the road 176．your dream. But the greatest one 177．from yourself.You need to decide 178．is the most important. Studying instead of watching TV will bring better exam grades, while saving twenty yuan instead of buying ice cream means you can buy a new book. As you get closer to your dream, it may 179．a little. This is goad as you have the chance to learn more kids and find new interests. So follow your 180．and you will surely have a better future.(19)change two sign will blind drop different coin but muchA blind boy sat on the steps of a building with a hat by his feet. He held up a sign which said: “I am blind, please help.” There were only a few 181．in the hat.A man was walking by. He took a few coins from his pocket and 182．them into the hat. He then took the sign, turned it around, and wrote some words. He put the sign back so that everyone who walked by 183．see the new words.Soon the hat began to fill up. A lot more people gave money to the 184．boy. That afternoon the man who had changed the 185．came to see how things were. The boy recognized his footsteps and asked, “Were you the one who 186．my sign this morning? What did you write?”The man said, "I only wrote the truth. I said what you said, 187．in a different way. I wrote: ‘Today is a beautiful day, but I cannot see it.’ ” Both signs told people that the boy was blind. But the first sign simply said the boy was blind. The 188．sign told people that they were so lucky that they were not blind. Should we be surprised that the second sign was 189．effective?Be thankful for what you have! But remember, be creative and learn to think 190．and positively.(20)miss, it, take, achieve, through, leave, where, result, change, differentAs a famous piano master, Paderewski was admired by many people. Wishing to make progress on the piano, one day, a boy was 191．to a Paderewski's concert by his mother. After they were seated, the mother saw a friend and walked up to greet her, 192．the boy alone in his seat.Catching the chance to explore the wonders of the concert hall, the little boy stood up and finally made his way 193．a door which was marked “No Admittance(进入)”. Soon, the hal l got dark and the concert was about to begin. When the mother returned to her seat, she discovered that the child was 194．.Suddenly, the lights focused on the piano on the stage. In surprise, the mother found her little boy sitting at the piano, without knowing 195．he was, the boy was picking out Twinkle. Twinkle. Little Star.At that moment, the great piano master appeared, moved to the piano quickly, and whispered(耳语)in the boys ear, “Don't stop. Keep playing.” Then he sat down and added a running obbl igato(伴奏). Together, the old master and the young beginner 196．an embarrassing situation into a wonderfully creative experience. 197．became one of the most exciting parts of the whole concert that evening. The people were touched so much by the master and the mother was shocked. She believed that such an experience would make a big 198．to her boy’s life.In fact, that's the way it is in life. What we can do on our own is hardly unusual. We tried our best, but the199．aren't exactly smooth music like what we have expected. But when we trust in the hands of a greater power, our life's work can be truly beautiful.Next time you set out to 200．great works, listen carefully. You can hear the voice of the master, whispering in your ear, “Don't stop. Keep playing.”(21)what, communicate, harm, child, ride, how, until, remember, way, bottle, recycle, ratherA whole school can take part in a recycling program and can set up an area where everyone can drop off things recyclable, such as some 201．and cans. Ink cartridges (墨水盒) are another things that can 202．. Often, these are thrown away which can be very 203．for the environment. Simply the way in which people within a school consume (消耗) everyday things can have an effect on the environment. Classes can save on paper by always 204．to write on both sides of each sheet; students should continue to use exercise books 205．each is complete, and teachers can put aside some paper, such as unneeded worksheets, for we use later 206．than waste it. Teachers and office workers should try to print less, and try to send e - mails as a form of 207．to reduce the amount of paper used if possible. Parents should consider 208．they take their kids to school, since using the car unnecessarily leads to the air pollution. Kids should be encouraged 209．bikes or walk to school if possible, or parents could offer lifts(搭车) to other 210．in the neighborhood to save on journeys.(22)on wide eat place as them situation call before useA school in England for children with autism（自闭症）is finding a new use for virtual reality（虚拟现实）, or VR headsets. Prior's Court is a school in Berkshire, southern England. The workers at the school are also 211．high technology to learn more about individual students.People with autism may find it hard to deal with 212．and situations they have not experienced213．. VR headsets make the wearer feel like they are in a different place. For example, someone wearing a VR headset can have a 360-degree view of a place 214．they turn around. With video, they can even hear the sounds of the place.Teachers at Prior's Court are using VR to introduce children to 215．like visiting a shopping mall or getting 216．a plane. They can do so in the safety of their classroom.The charity is also hoping data（数据）can help. 217．are trying a new data collection system. The system, 218．Prior Insight, puts together information about each young person's day: what they have 219．, how much sleep and exercise they have had and all their activities. Then it compares those facts to how they are behaving and medical events.“We're hoping to not only increase our knowledge and awareness about the world of young people with autism at Prior's Court, but we're also hoping to be able to, in time, share that with the 220．autism world. ”(23)old shoes stand feet pick small feel clear if useful young lookOne day, while I was sitting in my car in a parking area, I noticed a young boy in front of my car. “What is the boy doing?” I asked myself. “Oh, he is 221．up a coin from the ground.” When he 222．up, we saw each other 223．. He was not 224．than ten and was wearing one blue glove and one brown glove. His coat was too 225．for him.As he walked away, I saw his 226．, which were too old to wear. I rolled down my window and called him over.I asked him 227．he needed some m oney. He replied, “No. That’s Okay.”The parking area was wet. I could tell that his 228．were cold because he kept moving his weight from one foot to the other.“Please,” I held out a five-dollar bill(钞票), “it's not much, but when the money is shared, it is much more 229．.”He took off one of his gloves and took the money from my hand. His small hand was red and cold.He smiled and went away after saying “Thank you” to me. That smile made me 230．warm in the cold winter. I hope the boy felt warm too.(24)。

2021年中考总复习25分钟限时规范训练(含答案)限时训练3 生物的生存依赖一定的环境

2021年中考总复习25分钟限时规范训练(含答案)限时训练3 生物的生存依赖一定的环境

2021年中考总复习25分钟限时规范训练（含答案）姓名：班级：限时训练3生物的生存依赖一定的环境(时间25分钟分值25分)一、选择题(每小题1分，共15分)1．生物有许多区别于非生物的特征。

下列属于生物的是（A）A．地衣B．机器人C．钟乳石D．煮熟的鸡蛋2．《齐民要术》是我国古代的农业百科全书，对世界农业科学史产生了重要的影响。

其中对菜地的管理有“有草锄之”的论述。

种菜要锄草，因为杂草与蔬菜之间的关系是（A）A．竞争B．合作C．寄生D．共生3．“万物生长靠太阳”这句话体现出的生物与环境之间的关系是(C)A．生物适应环境B．生物影响环境C．环境影响生物D.生物与环境相互影响4．下列诗句中最能体现出光对植物生长影响的是AA．春色满园关不住，一枝红杏出墙来B．人间四月芳菲尽，山寺桃花始盛开C．竹外桃花三两枝，春江水暖鸭先知D．停车坐爱枫林晚，霜叶红于二月花5．“烟雨湿阑干，杏花惊蛰寒”，惊蛰节气前后气温转暖，雨水增多，杏花开放，惊雷惊醒了蛰伏于泥土中冬眠的昆虫。

这说明AA．环境影响生物B．环境适应生物C．生物影响环境D．生物适应环境6．下列选项中体现了生物适应环境的是AA．仙人球的叶变成刺状，以减少水分的散失B．千里之堤，溃于蚁穴C．过度放牧，破坏草场D．大树底下好乘凉7．地衣生活在岩石表面，可以从岩石中得到所需的营养物质；地衣又能够分泌地衣酸，对岩石有腐蚀作用。

这一事实说明(D)A．生物能适应环境B．生物不能适应环境，也不能影响环境C．生物能影响环境D．生物既能适应环境，又能影响环境8．生命世代相续，生生不息。

下列诗词中没有体现生殖现象的是BA．几处早莺争暖树，谁家新燕啄春泥B．碧玉妆成一树高，万条垂下绿丝绦C．穿花蛱蝶深深见，点水蜻蜓款款飞D．日出江花红胜火，春来江水绿如蓝9．下列所描述的生命现象与其实例不符的是CA．生物的生命活动需要营养——螳螂捕蝉，黄雀在后B．生物能对外界刺激作出反应——朵朵葵花向太阳C．生物能排出体内的代谢废物——蜻蜓点水D．生物能生长繁殖——离离原上草，一岁一枯荣10．谚语、俗语和古诗词中蕴含着一些生物学原理。

2021九年级数学中考总复习专题选择填空专项训练解题指导培优试题含答案解析

2021九年级数学中考总复习专题选择填空专项训练解题指导培优试题含答案解析

2021九年级数学中考总复习专题选择填空专项训练解题指导培优试题含答案解析1．如图，已知矩形ABCD沿着直线BD折叠，使点C落在C′处，BC′交AD于E，AD＝8，AB＝4，则DE的长为（）A．3B．4C．5D．62．如图，正方形纸片ABCD的边长为3，点E、F分别在边BC、CD上，将AB、AD分别沿AE、AF折叠，点B，D恰好都落在点G处，已知BE＝1，则EF的长为（）A．1.5B．2.5C．2.25D．33．如图，AB＝4，射线BM和AB互相垂直，点D是AB上的一个动点，点E在射线BM上，BE=12DB，作EF⊥DE，并截取EF＝DE，连接AF并延长交射线BM于点C，设BE＝x，BC＝y，则y关于x的函数解析式为．4．如图，在扇形OAB中，∠AOB＝110°，半径OA＝18，将扇形OAB沿着过点B的直线折叠，点O恰好落在AB̂上的点D处，折痕交OA于点C，则AD̂的长等于．（结果保留π）5．如图，在△ABC中，∠ABC＝60°，点P是△ABC内的一点，使得∠APB＝∠BPC＝∠CP A，且P A＝8，PC＝6，则PB＝．6．如图，在Rt△ABC中，∠ACB＝90°，∠B＝30°，AC＝2．作△ABC的高CD，作△CDB的高DC1，作△DC1B的高C1D1，…，如此下去，则得到的所有阴影三角形的面积之和为．7．如图所示，△ABC中，∠C＝90°，AC＝2，BC＝1，顶点A，C分别在x轴，y轴的正半轴上滑动，则点B到原点的最大距离是．8．如图，在平面直角坐标系xOy中，直线AB经过点A（﹣4，0）、B（0，4），⊙O的半径为1（O为坐标原点），点P在直线AB上，过点P作⊙O的一条切线PQ，Q为切点，则切线长PQ的最小值为．9．如图，已知一次函数y＝kx+b的图象经过点P（3，2），与反比例函数y=2x（x＞0）的图象交于点Q（m，n）．当一次函数y的值随x值的增大而增大时，m的取值范围是．10．如图，以扇形OAB的顶点O为原点，半径OB所在的直线为x轴，建立平面直角坐标系xOy，点B的坐标为（2，0），若抛物线y=12x2+k与扇形OAB的边界总有两个公共点，则实数k的取值范围是．中考总复习数学专题选择填空专项训练解题指导培优试题含答案解析一．试题（共11小题）1．如图，已知矩形ABCD 沿着直线BD 折叠，使点C 落在C ′处，BC ′交AD 于E ，AD ＝8，AB ＝4，则DE 的长为（ ）A ．3B ．4C ．5D ．6解：∵Rt △DC ′B 由Rt △DBC 翻折而成，∴CD ＝C ′D ＝AB ＝8，∠C ＝∠C ′＝90°，设DE ＝x ，则AE ＝8﹣x ，∵∠A ＝∠C ′＝90°，∠AEB ＝∠DEC ′，∴∠ABE ＝∠C ′DE ，在Rt △ABE 与Rt △C ′DE 中，{∠A =∠C′=90°AB =C′D ∠ABE =∠C′DE，∴Rt △ABE ≌Rt △C ′DE （ASA ），∴BE ＝DE ＝x ，在Rt △ABE 中，AB 2+AE 2＝BE 2，∴42+（8﹣x ）2＝x 2，解得：x ＝5，∴DE 的长为5．故选：C ．2．如图，正方形纸片ABCD 的边长为3，点E 、F 分别在边BC 、CD 上，将AB 、AD 分别沿AE 、AF 折叠，点B ，D 恰好都落在点G 处，已知BE ＝1，则EF 的长为（ ）A．1.5B．2.5C．2.25D．3解：∵正方形纸片ABCD的边长为3，∴∠C＝90°，BC＝CD＝3，根据折叠的性质得：EG＝BE＝1，GF＝DF，设DF＝x，则EF＝EG+GF＝1+x，FC＝DC﹣DF＝3﹣x，EC＝BC﹣BE＝3﹣1＝2，∵在Rt△EFC中，EF2＝EC2+FC2，即（x+1）2＝22+（3﹣x）2，解得：x＝1.5，∴DF＝1.5，EF＝1+1.5＝2.5．故选：B．3．如图，AB＝4，射线BM和AB互相垂直，点D是AB上的一个动点，点E在射线BM上，BE=12DB，作EF⊥DE，并截取EF＝DE，连接AF并延长交射线BM于点C，设BE＝x，BC＝y，则y关于x的函数解析式为y=12x4−x（0＜x≤2）．解：作FM⊥BC于M．∵∠DBE＝∠DEF＝∠EMF＝90°，∴∠DEB+∠BDE＝90°，∠DEB+∠FEM＝90°，∴∠BDE＝∠FEM．在△DBE和△EMF中，{∠BDE =∠FEM ∠B =∠EMF DE =EF，∴△DBE ≌△EMF ，∴FM ＝BE ＝x ，EM ＝BD ＝2BE ＝2x ，∵FM ∥AB ，∴FM AB =CM CB， ∴x 4=y−3x y， ∴y =12x 4−x （0＜x ≤2）．4．如图，在扇形OAB 中，∠AOB ＝110°，半径OA ＝18，将扇形OAB 沿着过点B 的直线折叠，点O 恰好落在AB ̂上的点D 处，折痕交OA 于点C ，则AD ̂的长等于 5π ．（结果保留π）解：连结OD ，∵△BCD 是由△BCO 翻折得到，∴∠CBD ＝∠CBO ，∠BOD ＝∠BDO ，∵OD ＝OB ，∴∠ODB ＝∠OBD ，∴∠ODB ＝2∠DBC ，∵∠ODB +∠DBC ＝90°，∴∠ODB ＝60°，∵OD ＝OB∴△ODB 是等边三角形，∴∠DOB ＝60°，∵∠AOB ＝110°，∴∠AOD ＝∠AOB ﹣∠DOB ＝50°，∴弧AD 的长=50π×18180=5π．故答案为：5π．5．如图，在△ABC 中，∠ABC ＝60°，点P 是△ABC 内的一点，使得∠APB ＝∠BPC ＝∠CP A ，且P A ＝8，PC ＝6，则PB ＝ 4√3 ．解：由题意∠APB ＝∠BPC ＝∠CP A ＝120°，设∠PBC ＝α，∠ABC ＝60°则∠ABP ＝60°﹣α，∴∠BAP ＝∠PBC ＝α，∴△ABP ∽△BCP ，∴AP BP =BP PC ，BP 2＝AP •PC ，∴BP =√AP ⋅PC =√48=4√3．故答案是：4√3．6．如图，在Rt △ABC 中，∠ACB ＝90°，∠B ＝30°，AC ＝2．作△ABC 的高CD ，作△CDB 的高DC 1，作△DC 1B 的高C 1D 1，…，如此下去，则得到的所有阴影三角形的面积之和为 6√37．解：∵DC 1∥AC ，∴Rt △ACD ∽△CDC 1，同理可证：Rt △C 1D 1D ∽Rt △C 1D 1C 2，…；即白色部分的小直角三角形与阴影部分的小直角三角形逐一对应相似，∵如图，在Rt △ABC 中，∠ACB ＝90°，∠B ＝30°，AC ＝2，∴AB ＝2AC ＝4，BC =√AB 2−AC 2=2√3．在Rt △ABC 中，CD ⊥AB ，由S =12AC •BC =12AB •CD ，故CD =√3，∴AC ：CD ＝2：√3，∴白色部分小直角三角形的面积和：阴影部分小直角三角形的面积和＝AC 2：CD 2＝4：3， 故S 阴影=37S △ABC =37×12×2×2√3=6√37．故答案是：6√37．7．如图所示，△ABC 中，∠C ＝90°，AC ＝2，BC ＝1，顶点A ，C 分别在x 轴，y 轴的正半轴上滑动，则点B 到原点的最大距离是 √2+1 ．解：设AC 的中点是D ，则OD =12AC ＝1，根据勾股定理得BD =√2，当B 、D 、O 在一条直线上时，点B 到原点O 的最大，最大距离是√2+1，故答案为：√2+1，8．如图，在平面直角坐标系xOy 中，直线AB 经过点A （﹣4，0）、B （0，4），⊙O 的半径为1（O为坐标原点），点P在直线AB上，过点P作⊙O的一条切线PQ，Q为切点，则切线长PQ的最小值为√7．解：连接OP、OQ．∵PQ是⊙O的切线，∴OQ⊥PQ；根据勾股定理知PQ2＝OP2﹣OQ2，∵当PO⊥AB时，线段PQ最短；又∵A（﹣4，0）、B（0，4），∴OA＝OB＝4，∴AB＝4√2∴OP=12AB＝2√2，∴PQ=√7；故答案为：√7．9．如图，已知一次函数y＝kx+b的图象经过点P（3，2），与反比例函数y=2x（x＞0）的图象交于点Q（m，n）．当一次函数y的值随x值的增大而增大时，m的取值范围是1＜m＜3．解：过点P 分别作y 轴与x 轴的垂线，分别交反比例函数图象于A 点和B 点，如图， 把y ＝2代入y =2x 得x ＝1；把x ＝3代入y =2x 得y =23，所以A 点坐标为（1，2），B 点坐标为（3，23）， 因为一次函数y 的值随x 值的增大而增大，所以Q 点只能在A 点与B 点之间，所以m 的取值范围是1＜m ＜3．故答案为1＜m ＜3．10．如图，以扇形OAB 的顶点O 为原点，半径OB 所在的直线为x 轴，建立平面直角坐标系xOy ，点B 的坐标为（2，0），若抛物线y =12x 2+k 与扇形OAB 的边界总有两个公共点，则实数k 的取值范围是 ﹣2＜k ＜12 ．解：由图可知，∠AOB ＝45°，∴直线OA 的解析式为y ＝x ，联立{y =xy =12x 2+k消掉y 得， x 2﹣2x +2k ＝0，第11页（共11页）△＝b 2﹣4ac ＝（﹣2）2﹣4×1×2k ＝0， 即k =12时，抛物线与OA 有一个交点， 此交点的横坐标为1，∵点B 的坐标为（2，0），∴OA ＝2，∴点A 的坐标为（√2，√2），∴交点在线段AO 上；当抛物线经过点B （2，0）时，12×4+k ＝0， 解得k ＝﹣2，∴要使抛物线y =12x 2+k 与扇形OAB 的边界总有两个公共点，实数k 的取值范围是﹣2＜k ＜12．故答案为：﹣2＜k ＜12。

江西省2021年九年级中考数学总复习选择填空提分特训(01)

江西省2021年九年级中考数学总复习选择填空提分特训(01)

选择填空提分特训(一)[限时:30分钟满分:36分]一、选择题(每小题3分,共18分)1.计算(-3)×5的结果是()A.-15B.15C.-2D.22.下列运算正确的是()A.-(x-y)2=-x2-2xy-y2B.a2+a2=a4C.a2·a3=a6D.(ay2)2=a2y43.我国平均每平方千米的土地一年从太阳得到的能量,相当于燃烧130000000 kg的煤所产生的能量.把130000000 kg用科学记数法可表示为()A.13×107 kgB.0.13×108 kgC.1.3×107 kgD.1.3×108 kg4.为了调查某小区居民的用水情况,随机抽查了若干户家庭的月用水量,结果如下表:月用水量(吨)3458户数2341则关于这若干户家庭的月用水量,下列说法错误的是()A.众数是4B.平均数是4.6C.调查了10户家庭的月用水量D.中位数是4.55.《九章算术》中,将两底面是直角三角形的棱柱称之为“堑堵”,已知某“堑堵”的三视图如图1所示,主视图中的虚线平分矩形的面积,则该“堑堵”的侧面积为()图1A.2B.4+2√2C.4+4√2D.6+4√26.如图2,点A是反比例函数y=m(m是常数,x>0)图象上的一个动点,过点A作x轴、y轴的平行线交反比例函数x(k为常数,k>0)图象于点B,C.当点A的横坐标逐渐增大时,△ABC的面积()y=kx图2A.先变大再变小B.先变小再变大C.不变D.无法判断二、填空题(每小题3分,共18分)7.因式分解:a2b-10ab+25b=.8.如图3,AB∥CD,EF⊥BD,垂足为F,∠1=43°,则∠2的度数为.图39.数学文化我国古代数学著作《增删算法统宗》记载“绳索量竿”问题:“一条竿子一条索,索比竿子长一托.折回索子却量竿,却比竿子短一托.”其大意为:现有一根竿和一条绳索,用绳索去量竿,绳索比竿长5尺;如果将绳索对折后再去量竿,就比竿短5尺.设绳索长x尺,竿长y尺,写出符合题意的方程组.10.如图4,点M,N分别是正五边形ABCDE的两边AB,BC上的点,且AM=BN,点O是正五边形的中心,则∠MON 的度数是度.图411.一元二次方程x2-3x+1=0的两个根为x1,x2,则x12+3x2+x1x2-2的值是.12.如图5,有一张三角形纸片ABC,∠A=80°,点D是AC边上一点,沿BD方向剪开三角形纸片后,发现所得两纸片均为等腰三角形,则∠C的度数可以是.图5【答案】1.A2.D3.D4.A5.C[解析]由主视图中虚线平分矩形的面积,可知俯视图是等腰直角三角形.如图所示,取AB的中点D,连接CD,则CD⊥AB.由左视图可知CD=1,∴AB=2CD=2,BC=AC=√2,该“堑堵”的侧面积为2×(2+2√2)=4+4√2,故选C.6.C[解析]设点A的坐标为(a,b),则点B kb ,b,点C a,ka,则S△ABC=12AB·AC=12a-kbb-ka=m2-k+k22m为定值.故选C.7.b(a-5)28.47°9.{x=y+5, 12x=y-510.72[解析]如图所示,连接OA,OB.∵OA=OB,∠OAM=∠OBN,AM=BN,∴△OAM≌△OBN.∴∠AOM=∠NOB,∴∠AOM+∠MOB=∠NOB+∠MOB,即∠AOB=∠MON.∵∠AOB是正五边形的中心角,∴∠MON=∠AOB=15×360°=72°.11.7[解析]因为x 2-3x+1=0的两个根为x 1,x 2,所以x 12=31,x 1+x 2=3,x 1x 2=1,所以x 12+3x 2+x 1x 2-2=31+3x 2+x 1x 2-2=3(x 1+x 2)+x 1x 2-3=7.12.25°或40°或10° [解析]分AB=AD 和AB=BD 和AD=BD 三种情况,利用等腰三角形的性质求出∠ADB ,再求出∠BDC ,然后根据等腰三角形两底角相等列式计算即可得解. 由题意知△ABD 与△DBC 均为等腰三角形. ①AB=BD ,此时∠ADB=∠A=80°, ∴∠BDC=180°-∠ADB=180°-80°=100°, ∴∠C=12×(180°-100°)=40°;②AB=AD ,此时∠ADB=12×(180°-∠A )=12×(180°-80°)=50°, ∴∠BDC=180°-∠ADB=180°-50°=130°, ∴∠C=12×(180°-130°)=25°;③AD=BD ,此时∠ADB=180°-2×80°=20°, ∴∠BDC=180°-∠ADB=180°-20°=160°, ∴∠C=12×(180°-160°)=10°.综上所述,∠C 的度数可以为25°或40°或10°. 故答案为25°或40°或10°.。

2021年内蒙古中考英语重点题型限时训练：完形填空+ 阅读理解(C)+ 选词填空+ 任务型阅读

2021年内蒙古中考英语重点题型限时训练：完形填空+ 阅读理解(C)+ 选词填空+ 任务型阅读

完形填空+ 阅读理解(C)+ 选词填空+ 任务型阅读[限时:30分钟]Ⅰ.完形填空A teacher decided to have her class play a game. The teacher told each student to bring a plastic bag with a few potatoes in it to school. Each student would “name” their potatoes after people they did not like. 1, the number of potatoes would be 2depending on how many people each student hated.The next day, with their potatoes, all the students went to school 3the game. Some kids had just one or two potatoes, while others had as many as five or six. The teacher then told the children that they would have to carry their potatoes with 4wherever they went for one week. Several days 5, some of the students started to complain(抱怨), as their potatoes began to rot(腐烂) and 6bad. The students who carried 7potatoes began to get unsatisfied with the heavy bags.One week later, the game 8. The teacher asked, “How did you 9carrying around your potatoes for a week?”The students complained once again. The teacher simply smiled and said, “This is what it’s like to carry hatred(仇恨) in your heart. You have to carry it with you wherever you go. If you can’t tolerate(容忍) carrying rotten potatoes for one week, how can you imagine having 10in your heart for a whole lifetime? Forgive others and move on with your life.”()1.A.However B.In a wordC.ThereforeD.After all()2.A.different B.similarC.the sameD.small()3.A.to watch B.to join inC.to winD.to lose()4.A.it B.thatC.themD.school bags()5.A.walked B.passed onC.passedD.walked by()6.A.taste B.smellC.soundD.feel()7.A.few B.someC.moreD.any()8.A.continued B.was doneC.beganD.was over()9.A.feel like B.look likeC.enjoyD.just like()10.A.love B.happinessC.friendshipD.hateⅠ. 阅读理解How can cactus(仙人掌) help the world? It can provide a new kind of sustainable fuel(可持续燃料). Nopalimex, a company in Mexico, has made one of the world’s first cactus-powered factories. The idea came to Sosa in Nopalimex when he was looking for ways to save money for his food factory. He now powers his factory with cactus gas(气体). The cost to produce the gas is half the cost of electricity.Soon, the gas produced by Nopalimex will also be used in cars. Nopalimex’s gas is made by using the prickly pear(刺梨)—a kind of cactus. To make the gas, the prickly pear is mixed together with manure(肥料). Then the mixture produces the gas that can be used as fuel.Why cactus? Fuel can be made from other plants such as corn, but there are a few reasons why cactus is a betterchoice. First, prickly pears need very little water and very little care, and grow well in hot temperatures. Today, these characteristics are key. Also, unlike corn, prickly pears don’t need space that could be used for other food plants. They can grow in areas where many food plants can’t grow. Last, after making gas from prickly pears and manure, their byproducts(副产品) can be used to help grow food.Scientists still need to do more research on making and using prickly pear gas. As projects like Nopalimex show, we have the possibility to produce more sustainable fuel to power our lives.1.How can cactus help the world?()A.By saving farmers’ money.B.By providing a new kind of sustainable fuel.C.By taking the place of corn.D.By producing a lot of electricity and gas.2.Why does Sosa power his factory with cactus gas?()A.Because he wants to save money for his food factory.B.Because he has grown a lot of cactus and corn in his town.C.Because he wants to be well-known around the world.D.Because he wants to grow cactus instead of corn.3.What does the underlined word “mixture” in Paragraph 2 refer to?()A.The Nopalimex company.B.The manure.C.The prickly pear and manure.D.Cactus.4.What is Paragraph 3 mainly about?()A.How cactus is used to produce gas.B.Where we can grow a lot of cactus.C.What Nopalimex does with cactus.D.Why cactus is a better choice to make fuel.Ⅰ. 选词填空阅读下列短文,用方框内所给单词的正确形式填空。

浙江省2021年中考数学复习选择填空限时练习汇总(批量下载)

浙江省2021年中考数学复习选择填空限时练习汇总(批量下载)

精品"正版〞资料系列,由本公司独创 .旨在将"人教版〞、〞苏教版"、〞北师大版"、〞华师大版"等涵盖几乎所有版本的教材教案、课件、导学案及同步练习和检测题分享给需要的朋友 .本资源创作于2021年8月,是当前最|新版本的教材资源 .包含本课对应内容,是您备课、上课、课后练习以及寒暑假预习的最|正确选择 .超级|资源 (共９套45页 )浙江省2021年（中|考）数学复习选择填空限时练习汇总选择填空限时练(一)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1. -2的相反数是( )A. B.-2 D. -2.如图X1 -1,下面几何体的俯视图是( )图X1 -1图X1 -23.[2021·绍兴]绿水青山就是金山银山,为了创造良好的生态生活环境,浙江省2021年清理河湖库塘淤泥约116000000方,数字116000000用科学记数法可以表示为 ( )×109 B.1.16×108C.1.16×107D.0.116×1094.把不等式组的解表示在数轴上,以下选项正确的选项是( )图X1 -35.用反证法证明命题:在一个三角形中,至|少有一个内角不大于60°.证明的第|一步是( )A.假设三个内角都不大于60°B.假设三个内角都大于60°C.假设三个内角至|多有一个大于60°D.假设三个内角至|多有两个大于60°6.从某市8所学校中抽取共1000名学生进行800米跑达标抽样检测,结果显示该市成绩达标的学生人数超过半数,达标率到达52.5%.如图X1 -4①、②反映的是本次抽样中的具体数据. 根据数据信息,以下判断:①小学高年级|被抽检人数为200人;②小学、初中、高中学生中,高中生800米跑达标率最|大;③小学生800米跑达标率低于33%;④高中生800米跑达标率超过70%.其中判断正确的有( )图X1 -4个个个个7.如图X1 -5,在▱ABCD中,用直尺和圆规作∠BAD的平分线AG交BC于点E,假设BF =6,AB =5,那么AE的长为( )图X1 -5B.68.关于x的方程ax +b =0(a≠0)的解为x = -2,点(1,3)是抛物线y =ax2 +bx +c(a≠0)上的一个点,那么以下四个点中一定在该抛物线上的是 ( )图X1 -6A.(2,3)B.(0,3)C.( -1,3)D.( -3,3)9.如图X1 -6,A,B是反比例函数y =(k>0,x>0)图象上的两点,BC∥x轴,交y轴于点C,动点P从坐标原点O出发,沿O→A→B→C匀速运动,终点为C,过运动路线上任意一点P作PM⊥x 轴于点M,PN⊥y轴于点N,设四边形OMPN的面积为S,P点运动的时间为t,那么S关于t的函数图象大致是( )图X1 -710.如图X1 -8,正方形ABCD的边长为6,点E,F分别在AB,AD上,假设CE =3,且∠ECF =45°,那么CF的长为( )图X1 -8C. D.二、填空题(每题4分,共24分)11.一组数据2,3,3,5,7的中位数是,方差是.12.如图X1 -9是一个斜体的 "土〞字,AB∥CD,∠1 =75°,那么∠2 = °.图X1 -913.为了了解某毕业班学生的睡眠时间情况,小红随机调查了该班15名同学,结果如下表:每天睡眠时间7 8 9 (单位:小时)人数 2 4 5 3 1 那么这15名同学每天睡眠时间的众数是小时,中位数是小时.14.如图X1 -10,将弧长为6π的扇形纸片AOB围成圆锥形纸帽,使扇形的两条半径OA与OB重合(粘连局部忽略不计),那么圆锥形纸帽的底面圆半径是.图X1 -1015.如图X1 -11,点B,D在反比例函数y =(a>0)的图象上,点A,C在反比例函数y =(b<0)的图象上,AB∥CD∥x轴,AB,CD在x轴的同侧,AB =4,CD =3,AB与CD间的距离为1,那么a -b的值是.图X1 -1116.如图X1 -12,点A(2,0),以OA为半径在第|一象限内作圆弧AB,使∠AOB =60°,点C为弧AB的中点,D为半径OA上一动点(不与点O,A重合),点A关于直线CD的对称点为E,假设点E落在半径OA上,那么点E的坐标为;假设点E落在半径OB上,那么点E的坐标为.图X1 -12|加加练|1.计算: +20210 -( -) -1+3tan30° +.2.解方程: + =3.3.先化简,再求值:2b2 +(a +b)(a -b) -(a -b)2,其中a = -3,b =.参考答案1.C2.A3.B4.B5.B6.C7.C8.D9.B10.A11.312.10513.8814.315.1216.(2 -2,0)( -1,3 -)加加练1.解:原式 =2 - +1 -( -3) +3× +2 =6 +2.2.解:去分母得x +( -2) =3(x -1),∴2x =1,∴x =.经检验,x =是原方程的解,∴原方程的解为x =.3.解:原式 =2b2 +a2 -b2 -(a2 -2ab +b2) =a2 +b2 -a2 +2ab -b2 =2ab.∵a = -3,b =,∴原式 =2×( -3)× = -3.选择填空限时练(二)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1.某小区经过改良用水设施,5年内小区居民累计节水39400吨,将39400用科学记数法表示为( )×104 B.3.94×104C.39.4×103D.4.0×1042.以下运算正确的选项是( )A.( -3)2 = -9B.( -1)2021×1 = -1C. -5 +3 =8D. -| -2| =23.以下图形中是轴对称图形但不是中|心对称图形的是 ( )A.等边三角形B.平行四边形C.矩形D.圆4.不等式3x<2(x +2)的解是( )A.x>2B.x<2C.x>4D.x<45.一组数据0, -1,1,2,3,那么这组数据的方差为( )B.1C.6.在Rt△ABC中,两直角边的长分别为6和8,那么其斜边上的中线长为( )B.37.在☉O中,圆心O到弦AB的距离为AB长度的一半,那么弦AB所对圆心角的大小为( )A.30°B.45°C.60°D.90°8.点C是线段AB的黄金分割点(AC>BC),那么以下结论正确的选项是 ( )2 =AC2 +BC22 =AC·BAC. =D. =9.如图X2 -1,D是等边三角形ABC边AB上的一点,且AD∶DB =1∶2,现将△ABC折叠,使点C 与D重合,折痕为EF,点E,F分别在AC和BC上,那么CE∶CF = ( )图X2 -1A. B. C. D.10.假设二次函数y =ax2 +bx +c(a≠0)的图象与x轴的交点坐标分别为(x1,0),(x2,0),且x1<x2,图象上有一点M(x0,y0)在x轴下方,对于以下说法:①b2-4ac>0;②x =x0是方程ax2 +bx +c =y0的解;③x1<x0<x2;④a(x0 -x1)(x0 -x2)<0.其中正确的选项是( )A.①③④B.①②④C.①②③D.②③二、填空题(每题4分,共24分)11.请写出一个解为x =1的一元一次方程: .12.计算:2tan60° +(2 -)0 -() -1 = .13.二次函数y =x2 +4x +5( -3≤x≤0)的最|大值是,最|小值是.14.当1<a<2时,代数式 +|1 -a| = .15.如图X2 -2,点A1,A2,…,A n均在直线y =x -1上,点B1,B2,…,B n均在双曲线y = -上,并且满足:A1B1⊥x轴,B1A2⊥y轴,A2B2⊥x轴,B2A3⊥y轴,…,A n B n⊥x轴,B n A n +1⊥y轴,…,记点A n 的横坐标为a n(n为正整数).假设a1 = -1,那么a3 = ,a2021 = .图X2 -216.如图X2 -3,在边长为2的菱形ABCD中,∠A =60°,M是AD边的中点,N是AB边上一动点,将△AMN沿MN所在的直线翻折得到△A'MN,连结A'C,那么A'C长度的最|小值是.图X2 -3参考答案1.B2.B3.A4.D5.D6.D7.D8.C9.A10.B11.x -1 =0(答案不唯一)12.2 -113.5 114.115.216. -1选择填空限时练(三)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1. -5的绝|对值等于( )B. -5C. D. -2.以下几何体中,俯视图为三角形的是( )图X3 -13.事件:在只装有2个红球和8个黑球的袋子里,摸出一个白球是( )A.可能事件B.随机事件C.不可能事件D.必然事件4.以下运算正确的选项是( )A.(2a2)3 =6a6B. -a2b2·3ab3 =-3a2b5C. + = -1D.· = -15.在一次中学生田径运动会上,参加男子跳高的20名运发动的成绩如下表:成绩/米人数 4 3 5 6 1 1 那么这些运发动成绩的众数为( )A.米B.米C.米D.米6.点( -2,y1),(3,y2)在一次函数y =2x -3的图象上,那么y1,y2,0的大小关系是( )1<y2<0 1<0<y22<0<y1 D.0<y1<y27.如图X3 -2,一架长米的梯子AB斜靠在墙上,梯子底端B到墙角C的距离为米,设梯子与地面所夹的锐角为α,那么cosα的值为( )图X3 -2A. B. C. D.8.我们知道方程组的解是现给出另一个方程组它的解是( )A. B.C. D.9.七巧板是我们祖先的一项卓越创造,被誉为 "东方魔板〞.如图X3 -3是一个七巧板迷宫,它恰好拼成了一个正方形ABCD,其中E,P分别是AD,CD的中点,一只蚂蚁从点A处沿图中实线爬行到出口点P处.假设AB =2,那么它爬行的最|短路程为( )图X3 -3A. B.1 +10.如图X3 -4,在▱ABCD中,∠DAB =60°,AB =10,AD =6,☉O分别切边AB,AD于点E,F,且圆心O恰好落在DE上.现将☉O沿AB方向滚动到与边BC相切(点O在▱ABCD的内部),那么圆心O移动的路径长为( )图X3 -4C.7 -D.10 -2二、填空题(每题4分,共24分)11.分解因式:ab +ac = .12.小红同学5月份各项消费情况的扇形统计图如图X3 -5,其中小红在学习用品上支出100元,那么在午餐上支出元.图X3 -513.如图X3 -6,在☉O中,C为优弧AB上一点,假设∠ACB =40°,那么∠AOB = 度.图X3 -614.甲、乙两工程队分别承接了250米,150米的道路铺设任务,乙比甲每天多铺设5米,甲完成铺设任务的时间是乙的2倍.设甲每天铺设x米,那么根据题意可列出方程: . 15.如图X3 -7,点A在第|一象限,作AB⊥x轴,垂足为点B,反比例函数y =的图象经过AB 的中点C,过点A作AD∥x轴,交该函数图象于点是AC的中点,连结OE,将△OBE沿直线OE对折到△OB'E,使OB'恰好经过点D,假设B'D =AE =1,那么k的值是.图X3 -716.如图X3 -8,矩形ABCD和正方形EFGH的中|心重合,AB =12,BC =16,EF =,分别延长FE,GF,HG和EH交AB,BC,CD,AD于点I,J,K,L.假设tan∠ALE =3,那么AI的长为,四边形AIEL的面积为.图X3 -8|加加练|1.计算:( -2021)0 + -9×.2.化简:(a +2)(a -2) -a(a +1).3.化简: +.参考答案1.A2.C3.C4.C5.C6.B7.A8.D9.B[解析] ∵正方形ABCD,E,P分别是AD,CD的中点,AB =2,∴AE =DE =DP =1,∠D =90°,∴EP = =,∴蚂蚁从点A沿图中实线爬到出口点P处,爬行的最|短路程为AE +EP =1 +.应选B.10.B[解析] 连结OA,OF.∵AB,AD分别与☉O相切于点E,F,∴OE⊥AB,OF⊥AD,∴∠OAE =∠OAD =30°.在Rt△ADE中,AD =6,∠ADE =30°,∴AE =AD =3,∴OE =AE· =.∵AD∥BC,∠DAB =60°,∴∠ABC =120°.设当运动停止时,☉O与BC,AB分别相切于点M,N,连结ON,OM,OB.那么∠BON =30°,且ON =,∴BN =ON·tan 30° =1,EN =AB -AE -BN =10-3-1 =6.∴圆心O移动的路径长为6.11.a(b +c)12.20013.8014. =15.12[解析] 如图,过D作DF⊥OB于F,设B'E与AD交于点G.∵AB⊥x轴,AD∥x轴,∴四边形ABFD是矩形,由折叠可得,∠B' =90° =∠A.又∵B'D =AE =1,∠DGB' =∠EGA,∴△DB'G≌△EAG,∴DG =EG,B'G =AG,∴AD =B'E =BE.又∵E是AC的中点,C是AB的中点,∴AE =CE =1,AC =BC =2,∴BE =3 =AD,AB =4 =DF.设C(a,2),那么D(a -3,4).∵反比例函数y =的图象经过点C,D,∴2a =4(a -3),解得a =6,∴C(6,2),∴k =6×2 =12.16.5[解析] 如图,过点E作EM⊥AB于点M,过点F作FN⊥AB于点N,过点E作EA1⊥AD于点A1,交FN于Q,过点G作GA2⊥AD,过点H作HP⊥A1E于P,∵tan∠1 =3,∴tan∠2 =3.又∵EF =,∴EQ =1,QF =3.∵矩形ABCD与正方形EFGH的中|心重合,∴AA1 =A2D =6,A1A2 =4 =PQ.同理得AN =8,NB =4,EM =6.易证△IME∽△EQF,∴ =,∴IM =2,∴IB =7,∴AI =5.∴A1E =7,∴A1L =,∴四边形AIEL的面积为 + =×(5 +7)×6 +×7× =.加加练1.解:原式 =1 +2 -9× =2.2.解:原式 =a2 -4 -a2 -a = -4 -a.3.解:原式 = = =a.选择填空限时练(四)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1.以下四个数: -1,0,,3.14,其中为无理数的是 ( )A. -1B.0C.2.以下计算正确的选项是( )3 +x4 =x73 -x4 =x -13·x4 =x73÷x4 =x3.如图X4 -1所示的支架的主视图是( )图X4 -1图X4 -24.如图X4 -3,电路图上有四个开关A,B,C,D和一个小灯泡,闭合开关D或同时闭合开关A,B,C 都可使小灯泡发光,那么任意闭合其中两个开关,小灯泡发光的概率是( )图X4 -3A. B.C. D.5.如图X4 -4,直线AB∥CD,∠GEB的平分线EF交CD于点F,∠1 =60°,那么∠2等于( )图X4 -4A.130°B.140°C.150°D.160°6.假设a -b =2ab,那么 -的值为( )A. -2B. -C.7.假设将直尺的0 cm刻度线与半径为5 cm的量角器的0°线对齐,并让量角器沿直尺的边缘无滑动地滚动(如图X4 -5),那么直尺上的10 cm刻度线对应量角器上的度数约为( )图X4 -5A.90°B.115°C.125°D.180°8.在某次体育测试中,九年级|一班女同学的一分钟仰卧起坐成绩(单位:个)如下表:成绩45 46 47 48 49 50 人数 1 2 4 2 5 1 这次测试成绩的中位数和众数分别为( )A.47,49B.48,49C.47.5,49D.48,509.如图X4 -6,在矩形ABCD中,AB =3,BC =5,点P是BC边上的一个动点(点P不与点B,C重合),现将△PCD沿直线PD折叠,使点C落到点C'处;作∠BPC'的平分线交AB于点E.设BP=x,BE =y,那么以下图象中,能表示y与x的函数关系的大致图象是( )图X4 -6图X4 -710.如图X4 -8,在平面直角坐标系中,直线l1⊥x轴于点A(2,0),点B是直线l1上的动点,直线l2:y =x +1交l1于点C,过点B作直线l3垂直于l2,垂足为D,过点O,B的直线l4交l2于点E.设直线l1,l2,l3围成的三角形的面积为S1,直线l2,l3,l4围成的三角形的面积为S2,且S2=S1,那么∠BOA的度数为 ( )图X4 -8A.15°B.30°C.15°或30°D.15°或75°二、填空题(每题4分,共24分)11.分解因式:a2 -4b2 = .12.二次根式中,x的取值范围是.13.如图X4 -9,把正三角形ABC的外接圆对折,使点A落在的中点F处,假设BC =6,那么折痕在△ABC内的局部DE的长为.图X4 -914.如图X4 -10,在边长为2的菱形ABCD中,∠AB C =120°,E,F分别为AD,CD上的动点,且AE +CF =2,那么线段EF长的最|小值是.图X4 -1015.如图X4 -11,一段抛物线:y = -x(x -3)(0≤x≤3),记为C1,它与x轴交于点O,A1;将C1绕点A1旋转180°得C2,交x轴于点A2;将C2绕点A2旋转180°得C3,交x轴于点A3;…;假设P(m,2)在第3段抛物线C3上,那么m = .图X4 -1116.对于两个不相等的实数a,b,我们规定符号max{a,b}表示a,b中较大的数,如max{2,4} =4.按照这个规定,方程max{x, -x} =的解为.|加加练|1.计算:( -)2 +| -4|×2 -1 -( -1)0.2.解不等式:3x -1≥2(x -1),并把它的解集在数轴上表示出来.图X4 -12 3.化简: +.参考答案1.C2.C3.D4.A5.C6.A7.B8.B9.D10.D11.(a +2b)(a -2b)12.x≤13.414.15.7或816.x =1 +或x = -1加加练1.解:原式 =3 +4× -1 =3 +2 -1 =4.2.解:去括号,得3x -1≥2x -2.移项、合并同类项,得x≥ -1.把不等式的解集在数轴上表示出来,如图:3.原式 = + = + =.选择填空限时练(五)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1.以下四个实数,2,0, -1,其中负数是 ( )A. B.2 D. -12.以下计算,结果等于a4的是( )A.a +3a 5 -a C.(a2)28÷a23.如图X5 -1所示,该圆柱体的左视图是 ( )图X5 -1图X5 -24.如图X5 -3,△ABC内接于☉O,∠A =68°,那么∠OBC等于( )图X5 -3A.22°B.26°C.32°D.34°5.某校数学兴趣小组在一次数学课外活动中,随机抽查了该校10名同学参加今年初中学业水平考试的体育成绩,统计结果如下表:成绩/分36 37 38 39 40 人数/人 1 2 1 4 2 表中表示成绩的数据中,中位数是( )分分分分6.用配方法解一元二次方程x2 -6x -10 =0,变形正确的选项是( )A.(x -3)2 =19B.(x +3)2 =19C.(x -3)2 =1D.(x +3)2 =17.不等式组的解集是( )≥2 B.1<x<2C.1<x≤2 ≤28.点( -1,y1),(1,0),(3,y2)都在一次函数y =kx -2的图象上,那么y1,y2,0的大小关系是( )A.0<y1<y21<0<y21<y2<0 2<0<y19.如图X5 -4,AB是半圆O的直径,半径OC⊥AB于点O,点D是的中点,连结CD,AD,OD,给出以下四个结论:①∠DOB =∠ADC;②CE =OE;③△ODE∽△ADO;④2CD2 =CE·AB.其中正确结论的序号是( )图X5 -4A.①③B.②④C.①④D.①②③10.如图X5 -5,矩形ABCD中,AB =8,BC =6,将矩形ABCD绕点A按逆时针方向旋转得到矩形AEFG,AE,FG分别交射线CD于点P,H,连结AH.假设P是CH的中点,那么△APH的周长为( )图X5 -5B.18二、填空题(每题4分,共24分)11.分解因式:a2-4a = .12.一个布袋里装有10个只有颜色不同的球,其中红球有m个,从布袋中随机摸出一个球记下颜色后放回、搅匀,再摸出一个球,通过大量重复试验后发现,摸到红球的频率稳定在左右,那么m的值为.13.某种品牌经过4,5月份连续两次降价,每部售价由5000元降到3600元.5月份降低的百分率是4月份降低的百分率的2倍,设4月份降低的百分率为x,根据题意可列方程: .14.如图X5 -6,用一个半径为60 cm,圆心角为150°的扇形围成一个圆锥,那么这个圆锥的底面半径为cm.图X5 -615.如图X5 -7,把菱形ABCD沿折痕AH翻折,使B点落在BC延长线上的点E处,连结DE.假设∠B =30°,那么∠CDE = °.图X5 -716.如图X5 -8,直角坐标系xOy中,直线y = -x +b分别交x,y轴的正半轴于点A,B,交反比例函数y = -的图象于点C,D(点C在第二象限内),过点C作CE⊥x轴于点E,记四边形OBCE 的面积为S1,△OBD的面积为S2,假设 =,那么CD的长为.图X5 -8|加加练|1.计算:( -2)0 -()2 +| -1|.2.解不等式组:3.解方程: -1 =.参考答案1.D2.C3.C4.A5.C6.A7.C8.B9.C10.C11.a(a -4)12.313.5000(1 -x)(1 -2x) =360014.2515.4516.5加加练1.解:原式 =1 -6 +1 = -4.2.解:解不等式①,得x> -3,解不等式②,得x<5,∴不等式组的解是 -3<x<5.3.解:原方程可化为2 -(x -2) =3x,解得x =1.经检验,x =1是原方程的解.所以原方程的解是x =1.选择填空限时练(六)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1.以下四个数中,是正整数的是 ( )A. -1C.2.据媒体报道,我国因环境污染造成的巨大经济损失,每年高达680000000元,这个数用科学记数法表示正确的选项是( )×109元 B.6.8×108元C.6.8×107元D.68×107元3.以下事件中,必然事件是 ( )A.今年夏季的雨量一定多B.下雨天每个人都打着伞C.二月份有30天D.我国冬季的平均气温比夏季的平均气温低4.如图X6 -1,点A,B,C,D,O都在方格纸的格点上,假设△COD是由△AOB绕点O按逆时针方向旋转而得,那么旋转的角度为( )图X6 -1A.30°B.45°C.90°D.135°5.一次函数y =2x -2的图象不经过的象限是 ( )A.第|一象限B.第二象限C.第三象限D.第四象限6.以下四个图形,其中是轴对称图形,且对称轴的条数为2的图形的个数是( )图X6 -2个个个个7.对于反比例函数y =,以下说法不正确的选项是( )A.点( -3, -1)在它的图象上B.它的图象在第|一,三象限随x的增大而减小D.当x>1时,y<38.如图X6 -3,在菱形ABCD中,对角线AC,BD相交于点O,E为BC的中点,那么以下式子一定成立的是( )图X6 -3A.AC =2OEB.BC =2OEC.AD =OED.OB =OE9.如图X6 -4,将长为2,宽为1的矩形纸片分割成n个三角形后,拼成面积为2的正方形,那么n≠( )图X6 -410.小阳在如图X6 -5①的扇形舞台上沿O➝M➝N匀速行走,他从点O出发,沿箭头所示的方向经过点M再走到点N,共用时70秒.有一台摄像机选择了一个固定的位置记录了小阳的走路过程,设小阳走路的时间为t(单位:秒),他与摄像机的距离为y(单位:米),表示y与t的函数关系的图象大致如图X6 -5②,那么这个固定位置可能是图X6 -5①中的( )图X6 -5A.点QB.点PC.点MD.点N二、填空题(每题4分,共24分)11.使代数式有意义的x的取值范围是.12.东山茶厂有甲、乙、丙三台包装机,同时分装质量为200克的茶叶.从它们各自分装的茶叶中分别随机抽取了15盒,测得它们的实际质量的方差如下表:甲包装机乙包装机丙包装机方差(克2)根据表中数据,三台包装机中, 包装机包装的茶叶质量最|稳定.13.如图X6 -6,l1是反比例函数y =在第|一象限内的图象,且过点A(2,1),l2与l1关于x轴对称,那么图象l2的函数解析式为(x>0).图X6 -614.将一个三角形经过放大后得到另一个三角形,如果所得三角形在原三角形的外部,这两个三角形各对应边平行且距离都相等,那么我们把这样的两个三角形叫做 "等距三角形〞,它们对应边之间的距离叫做 "等距〞.如果两个等边三角形是 "等距三角形〞,它们的 "等距〞是1,那么它们周长的差是.15.在平面直角坐标系内,以点P(1,2)为圆心,r为半径画圆,☉P与坐标轴恰好有三个交点,那么r的取值是.16.在平面直角坐标系xOy中,抛物线y = -x2 +2mx -m2 -m +1交y轴于点A,顶点为D,对称轴与x轴交于点H.(1)顶点D的坐标为(用含m的代数式表示);(2)当抛物线顶点D在第二象限时,如果∠ADH =∠AHO,那么m的值为.|加加练|1.计算:3-2-2cos60° +(12 -2006)0 -| -|.2.先化简,再求值:(1 -)÷,其中x请从 -2, -1,0,1,2中选一个恰当的数.参考答案1.D2.B3.D4.C5.B6.C7.C8.B9.A10.B11.x≥ -112.丙13.y = -14.615.或216.(1)(m,1 -m)(2)m = -1或m = -2加加练1.解:原式 = -2× +1 - = -.2.解:原式 =÷ =· =x +2.∵x≠0,1, -2,∴x可取 -1或2.当x =2时,原式 =2 +2 =4.(或当x = -1时,原式 = -1 +2 =1)选择填空限时练(七)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1.计算( -6) +5的结果是( )A. -11C. -12.函数y =中,自变量x的取值范围是( )≠2 ≥2C.x>2 ≥ -23.在以下 "绿色食品〞 "节能减排〞 "循环回收〞 "质量平安〞四个标志中,是轴对称图形的是( )图X7 -14.如图X7 -2是由4个相同的正方体搭成的几何体,那么其俯视图是 ( )图X7 -2图X7 -35.一个不透明的布袋中有2个白球,3个黑球,除颜色外其他都相同,从中随机摸出一个球,恰好为黑球的概率是( )A. B. C. D.6.如图X7 -4,矩形ABCD的两条对角线交于点O,假设∠AOD =120°,AB =6,那么AC等于( )图X7 -47.不等式2(x -1)≥x的解在数轴上表示为( )图X7 -58.如图X7 -6,D,E分别是△ABC的边AB,AC上的点,DE∥BC,且BD =3AD,那么AE∶AC等于( )图X7 -6∶3 ∶2∶3 ∶49.如图X7 -7,正方形ABCD的边长为1,分别以顶点A,B,C,D为圆心,1为半径画弧,四条弧交于点E,F,G,H,那么图中阴影局部的外围周长为( )图X7 -7A.πB.π D.π10.把三张大小相同的正方形卡片A,B,C叠放在一个底面为正方形的盒底上,底面未被卡片覆盖的局部用阴影表示.假设按图X7 -8①,②摆放,阴影局部的面积分别为S1和S2,那么S1和S2的大小关系是( )图X7 -81 =S2 1<S21>S2 D.无法确定二、填空题(每题4分,共24分)11.分解因式:ab-2a = .12.一组数据:2,1, -1,0,3,那么这组数据的中位数是.13.在同一平面直角坐标系内,将函数y =2x2 -3的图象向右平移2个单位,再向下平移1个单位后得到新图象的顶点坐标是.14.如图X7 -9,将Rt△ABC绕直角顶点A顺时针旋转90°,得到△AB'C',连结BB',假设∠1 =25°,那么∠C的度数是.图X7 -915.如图X7 -10,在平面直角坐标系中,直线y =kx +b与x轴,y轴分别交于点A(4,0),B(0,2),点C为线段AB上任意一点,过点C作CD⊥OA于点D,延长DC至|点E使CE =DC,作EF⊥y轴于点F,那么四边形ODEF的周长为.图X7 -1016.如图X7 -11,AB,CD是☉O的两条相互垂直的直径,E为半径OB上一点,且BE =3OE,延长CE交☉O于点F,线段AF与DO交于点M,那么的值是.图X7 -11|加加练|1.计算: -2cos 45° +() -1.2.化简: +.3.求满足不等式组的所有整数解.参考答案1.C2.B3.A4.A5.C6.C7.C8.D9.B10.A11.a(b -2)12.113.(2, -4)14.70°15.816.加加练1.解:原式 =2 -2× +2 = +2.2.解:原式 = = = =2.3.解:解x -3(x -2)≤8,得x≥ -1,解x -1<3 -x,得x<2,所以不等式组的解集为 -1≤x<2,其中所有的整数解为 -1,0,1.选择填空限时练(九)[限时:40分钟总分值:54分]一、选择题(每题3分,共30分)1.在 -2,0,,1这四个数中,最|大的数是( )A. -2B.0C.2.如图X9 -1是由几个相同的小正方体搭成的一个几何体,它的俯视图是 ( )图X9 -1图X9 -23.抽样调查某公司员工的年收入数据(单位:万元),结果如下表:年收入/万元 5 6 7 15 30 人数8 6 3 2 1 那么可以估计该公司员工中等年收入约为( )万元万元万元万元大型客机是中国具有自主知识产权的干线民用飞机,其零部件总数超过100万个,将100万用科学记数法表示为( )A.1×106B.100×104C.1×107D.0.1×1085.如图X9 -3,AB是☉O的弦,OC⊥AB,交☉O于点C,连结OA,OB,BC,假设∠ABC =20°,那么∠AOB的度数是( )图X9 -3A.40°B.50°C.70°D.80°6.不等式的解x≤2在数轴上表示为 ( )图X9 -47.如图X9 -5,在△ABC中,两条中线BE,CD相交于点O,那么S△DOE∶S△COB等于( )图X9 -5∶2 ∶3∶4 ∶38.小明进行两次定点投篮练习,第|一次a投b中(a≥b),第二次c投d中(c≥d),用新运算" 〞描述小明两次定点投篮总体命中率,那么以下算式合理的是( )A. =B. =C. =D. =9.如图X9 -6,抛物线y1 = -(x +2)2 -1与y2 =a(x -4)2 +3交于第四象限点A(1, -4),过点A作x轴的平行线,分别交两条抛物线于B,C两点,且D,E分别为顶点.那么以下结论正确的选项是( )图X9 -6A.AB<ACB.当x>1时,y1>y2C.△ACE是等边三角形D.△ABD是等腰三角形10.如图X9 -7,菱形ABCD中,∠ABC =60°,边长为3,P是对角线BD上的一个动点,那么BP +PC 的最|小值是 ( )图X9 -7A. B.D. +二、填空题(每题4分,共24分)11.分解因式:2m2 -8 = .12.如图X9 -8,把一张长方形纸带沿着直线GF折叠,假设∠CGF =30°,那么∠1的度数是.图X9 -813.某城市为了了解本市男女青少年平均身高发育情况,随机调查了6岁~18岁男女青少年各100人,制作成如图X9 -9所示的不同年龄平均身高统计图,从图中可知,该城市的男性青少年的身高高于同年龄女性的年龄段大概是.图X9 -914.如图X9 -10,P是边长为a的等边三角形ABC内任意一点,过点P分别作三角形三边的垂线PD,PE,PF,垂足分别点为D,E,F,那么图中阴影局部图形的面积总和为(用含a的式子表示) .图X9 -1015.如图X9 -11,正方形ABCD的边长为4,在这个正方形内作等边三角形EFG,使它们的中|心重合,那么△EFG的顶点到正方形ABCD的顶点的最|短距离是.图X9 -1116.下面是一种算法:输入任意一个数x,都是 "先乘2,再减去3”,进行第1次这样的运算,结果为y1,再对y1实施同样的运算,称为第2次运算,结果为y2,这样持续进行,要使第n次运算结果为0,即y n =0,那么最|初输入的数应该是.(用含有n的代数式表示)|加加练|1.化简:÷( -1).2.[2021·成都 ]为了美化环境,建设宜居成都,我市准备在一个广场上种植甲、乙两种花卉.经市场调查,甲种花卉的种植费用y(元)与种植面积x(m2)之间的函数关系如图X9 -12,乙种花卉的种植费用为每平方米100元.(1)直接写出当0≤x≤300和x>300时,y与x的函数关系式.(2)广场上甲、乙两种花卉种植面积共1200 m2,假设甲种花卉的种植面积不少于200 m2,且不超过乙种花卉种植面积的2倍,那么应该怎样分配甲、乙两种花卉的种植面积才能使种植总费用最|少?最|少费用为多少元?图X9 -12参考答案1.D2.D3.B4.A5.D6.B7.C8.C9.D10.B[解析] 如图,过点P作PM⊥AB于点M,过点C作CH⊥AB于点H.∵四边形ABCD是菱形,∠ABC =60°,∴∠PBM =∠ABC =30°,∴PM =PB,∴PB +PC =PC +PM.根据垂线段最|短可知,CP +PM的最|小值为CH的长.在Rt△CBH中,CH =BC·sin 60° =,∴PB +PC的最|小值为.应选B.11.2(m +2)(m -2)12.60°13.6~10岁和14~18岁14.15.4 -216.加加练1.解:原式 =÷=·=.2.解:(1)当0≤x≤300时,设函数关系式为y =k1x,过(300,39000),那么39000 =300k1,解得k1=130.∴当0≤x≤300时,y =130x;当x>300时,设函数关系式为y =k2x +b,过(300,39000)和(500,55000)两点,∴解得∴y =80x +15000.综上y =(2)设甲种花卉的种植面积为a m2,那么乙种花卉的种植面积为(1200 -a) m2.根据题意得解得200≤a≤800.当200≤a≤300时,总费用W1 =130a +100(1200 -a) =30a +120000,当a =200时,总费用最|少为W min =30×200 +120000 =126000(元);当300<a≤800时,总费用W2 =80a +15000 +100(1200 -a) = -20a +135000,当a =800时,总费用最|少为W min = -20×800 +135000 =119000(元).∵119000<126000,∴当a =800时,总费用最|少,为119000元,此时1200 -a =400.答:当甲、乙两种花卉种植面积分别为800 m2和400 m2时,种植总费用最|少,最|少费用为119000元.。

2021年广东中考数学选择填空题专练(3)(考试版)

2021年广东中考数学选择填空题专练(3)(考试版)

2021年广东中考数学选择填空题专练（三）（考试时间：25分钟；总分：58分)班级:___________姓名：___________座号：___________分数：___________一、单选题（每小题3分，共30分)1．的倒数是（）A、45B、54C、D、2．4月24日是中国航天日，1970年的这一天，我国自行设计、制造的第一颗人造地球卫星“东方红一号”成功发射，标志着中国从此进入了太空时代，它的运行轨道，距地球最近点439 000米．将439 000用科学记数法表示应为（）A．0.439×106 B．4.39×106 C．4.39×105 D．139×1033．下列倡导节约的图案中，是轴对称图形的是（）A．B．C．D．4．正十边形的外角和为（）A．180° B．360° C．720° D．1440°5．在数轴上，点A，B在原点O的两侧，分别表示数a，2，将点A向右平移1个单位长度，得到点C．若CO=BO，则a的值为（）A．-3B．-2C．-1D．16．下列运算正确的是（）A、3a2+4a2=7a4B、3a2﹣4a2=﹣a2C、3a•4a2=12a2D、7．下列等式不成立的是（）A、m2﹣16=（m﹣4）（m+4）B、m2+4m=m（m+4）C、m2﹣8m+16=（m﹣4）2D、m2+3m+9=（m+3）28．下列运算正确的是（）A、B、C、D、9．某校篮球班21名同学的身高如下表则该校蓝球班21名同学身高的众数和中位数分别是（单位：cm）（）A、186，186B、186，187C、186，188D、208，18810．如图，⊙O的弦AB垂直平分半径OC，若AB=√6，则⊙O的半径为（）A、√2B、2√2C、√22D、√62二、填空题（每小题4分，共28分)11．若分式1x x-的值为0，则x 的值为______. 12．如图，已知⊙ABC ，通过测量、计算得⊙ABC 的面积约为____cm 2.（结果保留一位小数）13．如图所示的网格是正方形网格，则PAB PBA ∠∠＋＝_____°（点A ，B ，P 是网格线交点）.14．在平面直角坐标系xOy 中，点A ()a b ，()00a b >>，在双曲线1k y x =上．点A 关于x 轴的对称点B 在双曲线2k y x=上，则12k k +的值为______. 15．把图1中的菱形沿对角线分成四个全等的直角三角形，将这四个直角三角形分别拼成如图2，图3所示的正方形，则图1中菱形的面积为______．16．小天想要计算一组数据92，90，94，86，99，85的方差20s ．在计算平均数的过程中，将这组数据中的每一个数都减去90，得到一组新数据2，0，4，-4，9，-5．记这组新数据的方差为21s ，则21s ______20s . (填“>”，“=”或“<”)17．在矩形ABCD 中，M ，N ，P ，Q 分别为边AB ，BC ，CD ，DA 上的点（不与端点重合）． 对于任意矩形ABCD ，下面四个结论中，①存在无数个四边形MNPQ 是平行四边形；②存在无数个四边形MNPQ 是矩形；③存在无数个四边形MNPQ 是菱形；④至少存在一个四边形MNPQ 是正方形．所有正确结论的序号是______．。

2021年浙江中考数学总复习选择填空提分特训(含答案)

2021年浙江中考数学总复习选择填空提分特训(含答案)

选择填空提分特训(二)[限时:40分钟满分:54分]一、选择题(每小题3分,共30分)1.计算23×(-2)的结果为()A.24B.-24C.22D.-222.使二次根式√x-1有意义的x的取值范围是()A.x≠1B.x>1C.x≤1D.x≥13.据统计,2019年十一国庆长假期间,南浔古镇共接待国内外游客约31.9万人次,与2018年同比增长16.43%.数据31.9万用科学记数法表示为()A.3.19×105B.3.19×106C.0.319×107D.319×1064.一个不透明布袋里装有1个白球、2个黑球、3个红球,它们除颜色外均相同.从中任意摸出一个球,则是红球的概率为()A.16B.13C.12D.235.如图X2-1所示的几何体的主视图为()图X2-1 图X2-26.下列说法错误的是()A.某商场对顾客健康码的审查,选择抽样调查B.在复学后,某校为了检查全校学生的体温,选择全面调查C.为了记录康复后的新冠肺炎病人的体温变化情况,适合选用折线统计图D.“发热病人的核酸检测呈阳性”是随机事件7.如图X2-3,已知圆锥的底面半径r为6 cm,高h为8 cm,则圆锥的侧面积为()图X2-3A.30π cm2B.48π cm2C.60π cm2D.80π cm28.《九章算术》是中国古代的数学专著,下面这道题是《九章算术》中第七章的一道题:“今有共买物,人出八,盈三;人出七,不足四,问人数、物价各几何?”译文:“几个人一起去购买某物品,如果每人出8钱,则多了3钱;如果每人出7钱,则少了4钱.问有多少人,物品的价格是多少?”设有x人,物品价格为y钱,可列方程组为()A.{8x-3=y7x+4=y B.{8x+3=y7x-4=y C.{y-8x=3y-7x=4D.{8x-y=37x-y=49.如图X2-4,把一个正方形三次对折后沿虚线剪下,则展开纸片“”得到的图形是()图X2-4图X2-510.如图X2-6,已知在△ABC中,∠C=90°,AC=6 cm,BC=8 cm,D是AC的中点,点P由点D出发,沿△ABC的各边顺时针方向运动,速度为7 cm/s,同时,点Q从C出发,沿△ABC的各边顺时针方向运动,速度为6 cm/s,当点P追上点Q时,两点停止运动.设运动时间为t(s),△DPQ的面积为S(cm2),则S关于t的函数图象大致为()图X2-6图X2-7二、填空题(每小题4分,共24分)11.实数-27的立方根是.12.分解因式:x2-xy=.13.计算:40°-15°30'=.14.如图X2-8,已知在▱ABCD中,AB=3.2,BC=2,以点C为圆心,适当长为半径画弧,交CD于P,交BC于点Q,再分PQ的长为半径画弧,两弧相交于点N,射线CN交DA的延长线于点E.则AE的长别以点P,Q为圆心,大于12是.图X2-8(x>0)的图象分别交矩形OABC的边AB,BC于点15.如图X2-9,已知在平面直角坐标系xOy中,反比例函数y=kxD ,E ,且BE=2CE ,若四边形ODBE 的面积为7,则k 的值为 .图X2-916.如图X2-10,已知Rt △ABD ≌Rt △BAC ,AD=3,AB=4,∠DAB=∠CBA=90°,点P 在这两个三角形的边上运动,若PA PB =13,则P A 的长为 .图X2-10附加训练17.计算:√8-2sin45°+(2-π)0-13-1.18.解分式方程:x -2x+2-1=16x 2-4.19.先化简,再求值:aa 2-ab ·(a 2-b 2),其中a=√2,b=-2√2.【参考答案】1.B2.D3.A4.C5.B6.A7.C8.A9.C 10.D 11.-3 12.x (x -y ) 13.24°30'(或24.5°) 14.1.2 15.3.516.1或√2或-2+3√65[解析] ∵Rt △ABD ≌Rt △BAC ,AD =3,AB =4,∴AC =BD =√AD 2+AB 2=5,当点P 在AB 边上时,∵PAPB =13,AB =4,∴P A =1; 当点P 在AD 边上时,∵PA PB=13,在Rt △P AB 中,P A 2+42=PB 2,∴P A 2+42=(3P A )2,解得P A =√2;当点P 在AC 边上时,过点P 作PE ⊥AB ,交AB 于点E ,连接PB ,PE =35AP ,AE =45AP ,BE =4-45AP .∵PA PB =13,∴PA√(35AP) +(4-45AP) =13,∴5P A 2+4P A -10=0,解得P A =-2-3√65(舍去)或P A =-2+3√65.故P A 的长为1或√2或-2+3√65.17.解:原式=2√2-2×√22+1-3=2√2−√2+1-3=√2-2.18.解:最简公分母为(x +2)(x -2),去分母得:(x -2)2-(x +2)(x -2)=16,整理得-4x +8=16,解得x =-2,经检验x =-2是原方程的增根,因此原方程无解. 19.解:aa 2-ab ·(a 2-b 2)=aa (a -b )·(a +b )(a -b )=a +b. 当a =√2,b =-2√2时,原式=√2+(-2√2)=-√2.。

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2021中考选择填空满分限时训31．（4分）计算1﹣3的结果是（）A．2B．﹣2C．4D．﹣42．（4分）用三个相同的正方体搭成如图所示的立体图形，则该立体图形的主视图是（）A．B．C．D．3．（4分）计算2a2•3a4的结果是（）A．5a6B．5a8C．6a6D．6a84．（4分）无理数√10在（）A．2和3之间B．3和4之间C．4和5之间D．5和6之间5．（4分）在一次数学测试中，小明成绩72分，超过班级半数同学的成绩，分折得出这个结论所用的统计量是（）A．中位数B．众数C．平均数D．方差6．（4分）如图，把△ABC先向右平移3个单位，再向上平移2个单位得到△DEF，则顶点C（0，﹣1）对应点的坐标为（）A．（0，0）B．（1，2）C．（1，3）D．（3，1）7．（4分）如图，已知线段AB ，分别以A ，B 为圆心，大于12AB 同样长为半径画弧，两弧交于点C ，D ，连接AC ，AD ，BC ，BD ，CD ，则下列说法错误的是（ ）A ．AB 平分∠CADB ．CD 平分∠ACBC ．AB ⊥CDD ．AB ＝CD8．（4分）下列是关于某个四边形的三个结论：①它的对角线相等；②它是一个正方形；③它是一个矩形．下列推理过程正确的是（ ） A ．由②推出③，由③推出① B ．由①推出②，由②推出③C ．由③推出①，由①推出②D ．由①推出③，由③推出②9．（4分）如图1，小球从左侧的斜坡滚下，到达底端后又沿着右侧斜坡向上滚，在这个过程中，小球的运动速度v （单位：m /s ）与运动时间t （单位：s ）的函数图象如图2，则该小球的运动路程y （单位：m ）与运动时间t （单位：s ）之间的函数图象大致是（ ）A ．B ．C ．D ．10．（4分）把一张宽为1cm 的长方形纸片ABCD 折叠成如图所示的阴影图案，顶点A ，D 互相重合，中间空白部分是以E 为直角顶点，腰长为2cm 的等腰直角三角形，则纸片的长AD （单位：cm ）为（ ）A ．7+3√2B ．7+4√2C ．8+3√2D ．8+4√2二、填空题（本题有6小题，每小题5分，共30分） 11．（5分）因式分解：x 2﹣9＝ ． 12．（5分）计算1x −13x的结果是 ．13．（5分）如图，等边三角形纸片ABC 的边长为6，E ，F 是边BC 上的三等分点．分别过点E ，F 沿着平行于BA ，CA 方向各剪一刀，则剪下的△DEF 的周长是 ．14．（5分）甲、乙两位同学在10次定点投篮训练中（每次训练投8个），各次训练成绩（投中个数）的折线统计图如图所示，他们成绩的方差分别为s 甲2与S 乙2，则s 甲2 S乙2．（填“＞”、“＝”、“＜“中的一个）15．（5分）如图，在△ABC中，D是边BC上的一点，以AD为直径的⊙O交AC于点E，连接DE．若⊙O与BC相切，∠ADE＝55°，则∠C的度数为．16．（5分）用四块大正方形地砖和一块小正方形地砖拼成如图所示的实线图案，每块大正方形地砖面积为a，小正方形地砖面积为b，依次连接四块大正方形地砖的中心得到正方形ABCD．则正方形ABCD的面积为．（用含a，b的代数式表示）17．（8分）计算：|﹣3|+√8−√2．参考答案与试题解析1．（4分）计算1﹣3的结果是（）A．2B．﹣2C．4D．﹣4【分析】根据有理数的加减法法则计算即可判断．【解答】解：1﹣3＝1+（﹣3）＝﹣2．故选：B．【点评】本题主要考查了有理数的减法法则，减去一个数，等于加上这个数的相反数．2．（4分）用三个相同的正方体搭成如图所示的立体图形，则该立体图形的主视图是（）A．B．C．D．【分析】从正面看所得到的图形即为主视图，因此选项A的图形符合题意．【解答】解：根据主视图的意义可知，选项A符合题意，故选：A．【点评】考查简单几何体的三视图的画法，从不同方向对问题进行正投影所得到的图形分别为主视图、左视图、俯视图．3．（4分）计算2a2•3a4的结果是（）A．5a6B．5a8C．6a6D．6a8【分析】直接利用单项式乘单项式运算法则计算得出答案．【解答】解：2a2•3a4＝6a6．故选：C．【点评】此题主要考查了单项式乘单项式，正确掌握相关运算法则是解题关键．4．（4分）无理数√10在（）A．2和3之间B．3和4之间C．4和5之间D．5和6之间【分析】由√9＜√10＜√16可以得到答案．【解答】解：∵3＜√10＜4，故选：B．【点评】此题考查了估算无理数的大小，熟练掌握估算无理数的方法是解本题的关键．5．（4分）在一次数学测试中，小明成绩72分，超过班级半数同学的成绩，分折得出这个结论所用的统计量是（）A．中位数B．众数C．平均数D．方差【分析】根据中位数的意义求解可得．【解答】解：班级数学成绩排列后，最中间一个数或最中间两个分数的平均数是这组成绩的中位数，半数同学的成绩位于中位数或中位数以下，小明成绩超过班级半数同学的成绩所用的统计量是中位数，故选：A．【点评】本题主要考查统计量的选择，解题的关键是掌握中位数、众数、平均数及方差的定义和意义．6．（4分）如图，把△ABC 先向右平移3个单位，再向上平移2个单位得到△DEF ，则顶点C （0，﹣1）对应点的坐标为（ ）A ．（0，0）B ．（1，2）C ．（1，3）D ．（3，1）【分析】利用平移规律进而得出答案．【解答】解：∵把△ABC 先向右平移3个单位，再向上平移2个单位得到△DEF ，顶点C （0，﹣1）， ∴C （0+3，﹣1+2）， 即C （3，1）， 故选：D ．【点评】此题主要考查了坐标与图形变化﹣平移，正确得出对应点位置是解题关键． 7．（4分）如图，已知线段AB ，分别以A ，B 为圆心，大于12AB 同样长为半径画弧，两弧交于点C ，D ，连接AC ，AD ，BC ，BD ，CD ，则下列说法错误的是（ ）A ．AB 平分∠CADB ．CD 平分∠ACBC ．AB ⊥CDD ．AB ＝CD【分析】根据作图判断出四边形ACBD 是菱形，再根据菱形的性质：菱形的对角线平分一组对角、菱形的对角线互相垂直平分可得出答案． 【解答】解：由作图知AC ＝AD ＝BC ＝BD ， ∴四边形ACBD 是菱形，∴AB 平分∠CAD 、CD 平分∠ACB 、AB ⊥CD ，不能判断AB＝CD，故选：D．【点评】本题主要考查作图﹣基本作图，解题的关键是掌握菱形的判定与性质．8．（4分）下列是关于某个四边形的三个结论：①它的对角线相等；②它是一个正方形；③它是一个矩形．下列推理过程正确的是（）A．由②推出③，由③推出①B．由①推出②，由②推出③C．由③推出①，由①推出②D．由①推出③，由③推出②【分析】根据对角线相等的四边形推不出是正方形或矩形即可判断．【解答】解：对角线相等的四边形推不出是正方形或矩形，故①→②，①→③错误，故选项B，C，D错误，故选：A．【点评】本题考查正方形的判定和性质，矩形的判定等知识，解题的关键是灵活运用所学知识解决问题，属于中考常考题型．9．（4分）如图1，小球从左侧的斜坡滚下，到达底端后又沿着右侧斜坡向上滚，在这个过程中，小球的运动速度v（单位：m/s）与运动时间t（单位：s）的函数图象如图2，则该小球的运动路程y（单位：m）与运动时间t（单位：s）之间的函数图象大致是（）A．B．C．D．【分析】小球从左侧的斜坡滚下是匀变速运动，运动的路程y是t的二次函数，图象是先缓后陡，由此即可判断．【解答】解：小球从左侧的斜坡滚下是匀变速运动，运动的路程y是t的二次函数，图象是先缓后陡，在右侧上升时，情形与左侧相反，故选：C．【点评】本题考查动点问题函数图象，解题的关键是理解题意，灵活运用所学知识解决问题．10．（4分）把一张宽为1cm的长方形纸片ABCD折叠成如图所示的阴影图案，顶点A，D 互相重合，中间空白部分是以E为直角顶点，腰长为2cm的等腰直角三角形，则纸片的长AD（单位：cm）为（）A．7+3√2B．7+4√2C．8+3√2D．8+4√2【分析】如图，过点M作MH⊥A′R于H，过点N作NJ⊥A′W于J．想办法求出AR，RM，MN，NW，WD即可解决问题．【解答】解：如图，过点M作MH⊥A′R于H，过点N作NJ⊥A′W于J．由题意△EMN是等腰直角三角形，EM＝EN＝2，MN＝2√2，∵四边形EMHK是矩形，∴EK＝A′K＝MH＝1，KH＝EM＝2，∵△RMH是等腰直角三角形，∴RH＝MH＝1，RM=√2，同法可证NW=√2，由题意AR＝RA′＝A′W＝WD＝4，∴AD＝AR+RM+MN+NW+DW＝4+√2+2√2+√2+4＝8+4√2，故选：D．【点评】本题考查翻折变换，等腰直角三角形的判定和性质，矩形的性质等知识，解题的关键是学会添加常用辅助线，构造特殊三角形或特殊四边形解决问题．11．（5分）因式分解：x 2﹣9＝ （x +3）（x ﹣3） ． 【分析】原式利用平方差公式分解即可． 【解答】解：原式＝（x +3）（x ﹣3）， 故答案为：（x +3）（x ﹣3）．【点评】此题考查了因式分解﹣运用公式法，熟练掌握平方差公式是解本题的关键． 12．（5分）计算1x −13x的结果是23x．【分析】先通分，再相减即可求解． 【解答】解：1x −13x=33x−13x=23x．故答案为：23x．【点评】考查了分式加减法，把分母不相同的几个分式化成分母相同的分式，叫做通分，经过通分，异分母分式的加减就转化为同分母分式的加减．13．（5分）如图，等边三角形纸片ABC 的边长为6，E ，F 是边BC 上的三等分点．分别过点E ，F 沿着平行于BA ，CA 方向各剪一刀，则剪下的△DEF 的周长是 6 ．【分析】根据三等分点的定义可求EF 的长，再根据等边三角形的判定与性质即可求解． 【解答】解：∵等边三角形纸片ABC 的边长为6，E ，F 是边BC 上的三等分点， ∴EF ＝2，∵DE ∥AB ，DF ∥AC ， ∴△DEF 是等边三角形，∴剪下的△DEF 的周长是2×3＝6． 故答案为：6．【点评】考查了等边三角形的性质，平行线的性质，关键是证明△DEF 是等边三角形． 14．（5分）甲、乙两位同学在10次定点投篮训练中（每次训练投8个），各次训练成绩（投中个数）的折线统计图如图所示，他们成绩的方差分别为s 甲2与S 乙2，则s 甲2 ＜ S 乙2．（填“＞”、“＝”、“＜“中的一个）【分析】利用折线统计图可判断乙同学的成绩波动较大，然后根据方差的意义可得到甲、乙的方差的大小．【解答】解：由折线统计图得乙同学的成绩波动较大，所以s甲2＜S乙2．故答案为：＜．【点评】本题考查了条形统计图：条形统计图是用线段长度表示数据，根据数量的多少画成长短不同的矩形直条，然后按顺序把这些直条排列起来．也考查了方差的意义．15．（5分）如图，在△ABC中，D是边BC上的一点，以AD为直径的⊙O交AC于点E，连接DE．若⊙O与BC相切，∠ADE＝55°，则∠C的度数为55°．【分析】由直径所对的圆周角为直角得∠AED＝90°，由切线的性质可得∠ADC＝90°，然后由同角的余角相等可得∠C＝∠ADE＝55°．【解答】解：∵AD为⊙O的直径，∴∠AED＝90°，∴∠ADE+∠DAE＝90°；∵⊙O与BC相切，∴∠ADC＝90°，∴∠C+∠DAE＝90°，∴∠C＝∠ADE，∵∠ADE＝55°，∴∠C＝55°．故答案为：55°．【点评】本题考查了切线的性质、圆的相关概念及性质及互余关系等知识点，熟练掌握圆的相关性质是解题的关键．16．（5分）用四块大正方形地砖和一块小正方形地砖拼成如图所示的实线图案，每块大正方形地砖面积为a，小正方形地砖面积为b，依次连接四块大正方形地砖的中心得到正方形ABCD．则正方形ABCD的面积为a+b．（用含a，b的代数式表示）【分析】如图，正方形ABCD是由4个直角三角形和一个小正方形组成，4个直角三角形的面积和等于大正方形的面积a，由此即可解决问题．【解答】解：如图，正方形ABCD是由4个直角三角形和一个小正方形组成，4个直角三角形的面积和等于大正方形的面积a．故正方形ABCD的面积＝a+b．故答案为a+b．【点评】本题考查中心对称，全等三角形的判定和性质，图形的拼剪等知识，解题的关键灵活运用所学知识解决问题，属于中考常考题型．17．（8分）计算：|﹣3|+√8−√2．【分析】直接利用绝对值的性质和二次根式的性质化简得出答案．【解答】解：原式＝3+2√2−√2＝3+√2．【点评】此题主要考查了实数运算，正确化简二次根式是解题关键．11。

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