2020-2021学年广东省深圳市宝安区新安中学(集团)外国语学校七上月考数学试卷

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2020-2021学年广东省深圳市外国语学校七年级上期中数学测试卷(解析版)

2020-2021学年广东省深圳市外国语学校七年级上期中数学测试卷(解析版)

2020-2021学年广东省深圳市外国语学校七上期中数学测试卷一、选择题(共10小题;共50分)1. 下列算式中,运算结果为负数的是A. B. C. D.2. 年“五一”假期期间,某市共接待国内、外游客万人次,实现旅游综合收入亿元,则“旅游综合收入”用科学记数法表示正确的是A. 元B. 元C. 元D. 元3. 如图所示是一个小正方体的展开图,把展开图折叠成小正方体,有“粤”字一面的相对面上的字是A. 澳B. 大C. 湾D. 区4. 汛期的某一天,某水库上午时的水位是,随后水位以每小时的速度上涨,中午时开始开闸泄洪,之后水位以每小时的速度下降,则当天下午时,该水库的水位是A. B. C. D.5. 已知和是同类项,则的值为A. B. C. D.6. 若的相反数是,,且,则的值是A. B. 或 C. 或 D.7. 如图,几何体的底层由四个大小相同的小立方块组成,则从左面看到的图形是A. B.C. D.8. 用一根长为(单位:)的铁丝,首尾相接围成一个正方形,要将它按如图的方式向外等距扩(单位:)得到新的正方形,则这根铁丝需增加A. B. C. D.9. 下列图形都是由同样大小的小圆圈按一定规律所组成的,其中第①个图形中一共有个小圆圈,第②个图形中一共有个小圆圈,第③个图形中一共有个小圆圈,,按此规律排列,则第个图形中小圆圈的个数为A. B. C. D.10. 定义一种对正整数的“ ”运算,①当为奇数时,结果为;②当为偶数时,结果为(其中是使为奇数的正整数),并且运算重复进行,例如,取,则当,则第次“ ”运算的结果是A. B. C. D.二、填空题(共8小题;共40分)11. 在代数式,,,,,中,整式有个,是单项式,是多项式.12. 如果用平面截掉一个长方体的一个角(即切去一个三棱锥),则剩下的几何体最多有个顶点,最少有条棱.13. 若一个棱柱共有条棱,则它的侧面数是.14. 检查个篮球的质量,把超过标准质量的克数记作正数,不足的克数记作负数,检查结果如表:(1)最接近标准质量的是号篮球;(2)质量最大的篮球比质量最小的篮球重.15. 如图,是一个简单的数值计算程序,当输入的的值为,则输出的结果为.16. 有理数,,,,在数轴上的位置如图,则.17. 若多项式是三次三项式,则代数式的值是.18. 探索与发现:下面是用分数(数字表示面积)砌成的“分数墙”,则整面“分数墙”的总面积是.三、解答题(共8小题;共104分)19. 计算:(1);(2);(3);(4).20. 将,,,在数轴上表示出来,并用“ ”号连接.21. 由个棱长为的相同小立方块搭成的几何体如图所示.(1)请画出它从正面、左面和上面看到的形状图;(2)请计算它的表面积.22. 如图,若图中平面展开图折叠成正方体后,相对面上的两个数字之和为,求的值.23. 先化简再求值:(1),其中,,;(2),其中,.24. 某自行车厂计划平均每天生产辆自行车,但由于各种原因,实际每天生产量与计划量相比有出入,下表是某周的生产情况(超产记为正,减产记为负):(1)根据记录的数据求该厂星期五生产自行车的辆数;(2)根据记录的数据求该厂本周实际生产自行车的辆数;(3)该厂实行每天计件工资制,每生产一辆自行车可得元,若超额完成任务,则超过部分每辆在元基础上另奖元;少生产一辆扣元.那么该厂工人这一周的工资总额是多少元?(4)若将()中的“实行每天计件工资制”改为“实行每周计件工资制”,其他条件不变,在此方式下这一周工人的工资与按日计件的工资哪一个较高?请说明理由.25. 小梅将边长分别为,,,,,,的若干个正方形按一定规律拼成不同的长方形,如图所示.(1)求第四个长方形的周长;(2)当时,求第五个长方形的面积.(用科学记数法表示)26. 如图所示,将连续的奇数,,,,排列成如下的数表,用十字形框框出个数.(1)探究规律一:设十字框中间的奇数为,则框中五个奇数的和用含的整式表示为,这说明被十字框框中的五个奇数的和一定是正整数的倍数,这个正整数是;(2)探究规律二:落在十字框中间且位于第二列的一组奇数是,,,,则这一组数可以用整式表示为(为正整数),同样,落在十字框中间且位于第三列的一组奇数可以表示为.(用含的式子表示)(3)运用规律()被十字框框中的五个奇数的和可能是吗?若能,请求出这五个数,若不能,请说明理由;()请问()中的十字框中间的奇数落在第几行第几列?答案第一部分1. B2. C3. B4. B5. C6. A7. B 【解析】从左面看该几何体,看得到的图形下面有两个小正方形,上面左侧有一个小正方形.8. B9. D 【解析】因为第个图形有个圆圈,第个图形有个圆圈,第个图形有个圆圈,所以第个图形有个圆圈.10. D第二部分11. ,,,,【解析】其中的,,,是整式,,是单项式,,是多项式.12. ,13.【解析】由棱柱的特点可知,这是一个五棱柱.故它有个侧面.14. ,【解析】(1)因为,,,,,,所以号球质量接近标准质量;(2)质量最大的篮球比质量最小的篮球重:(克).15.【解析】把代入得,把代入得,则输出的结果为.16.【解析】由数轴可得,,,,,则.17. 或18.第三部分19. (1)(2)(3)(4)20. ,,,,如图所示.故 .21. (1)如图所示.(2)从正面看,有个面,从后面看有个面,从上面看,有个面,从下面看,有个面,从左面看,有个面,从右面看,有个面,中间空处的两边两个正方形,有个面,所以表面积为.22. 由题意得与相对的是,所以,,与相对的是.所以,,与相对的是,所以,,所以.23. (1)当,,时,(2)当,时,24. (1)(辆).即该厂星期五生产自行车辆.(2)即该厂本周实际生产自行车辆.(3)即该厂工人这一周的工资总额是元.(4)实行每周计件工资制时工资较高.理由如下:实行每周计件工资制的工资为.故按周计件工资制的一周工资较高.25. (1)第一个的周长为,第二个的周长为,第三个的周长为,第四个的周长为,即第四个长方形的周长为.(2)由()可推出第个长方形的宽为第个长方形的长,第个长方形的长为第个长方形的长和宽的和.可得第五个长方形的宽为,长为,当时,第五个长方形的面积为.26. (1);【解析】设十字框中间的奇数为,则框中五个奇数中其他四个数分别为,,,.所以框中五个奇数的和为,所以这个正整数是.(2)【解析】因为落在十字框中间且位于第二列的一组数为,所以落在十字框中间且位于第三列的一组数为.(3)()根据题意得,则,其他四个数为,,,.所以这五个数分别为,,,,.()因为,所以为该数表的第个数又因为,所以()中的十字框中间的奇数落在第行第列.。

广东省深圳市宝安中学2020-2021学年七年级上学期10月月考英语试题

广东省深圳市宝安中学2020-2021学年七年级上学期10月月考英语试题

广东省深圳市宝安中学2020-2021学年七年级上学期10月月考英语试题学校:___________姓名:___________班级:___________考号:___________一、单项选择1.—When will our meeting end?—Maybe in about two hours.A.start B.finish C.take2.— Excuse me, is there a supermarket close to the hospital ?—Yes, there is. It’s over there.A.not near B.not far away from C.between 3.—Everyone should be at the school gate at 8:30 tomorrow morning. Are you clear?—Yes. We won’t be late.A.some people B.all people C.no people4.— I can’t play the piano very well.—Well, you should spend more time on it and do more practice.A.sport B.homework C.exercise5.—How was your trip to Beijing last month?—Very good. I had a good time there.A.enjoyed ourselves B.had fun C.had free time6.— Do you know Peter?—Yes, he is my classmate. He is friendly to everyone.A.important B.kind C.own7.—South China Tigers are becoming fewer and fewer.—We must do something to protect them.A.keep…clean B.keep…quiet C.keep…safe8.— He is such a good man.—Yes. He often provides some food and clothes to help the poor people.A.gives B.buys C.puts9.— What does your sister ________?—She has long hair and brown eyes.A.be like B.good at C.look like10.— Jane, what are you doing near the mailbox?— I’m looking for a letter, because I want to ________ my cousin.A.take part in B.make friends with C.hear from11.— _________ pollution is becoming more and more serious.—Yes. So we should not throw rubbish on the ground.A.Land B.Air C.Water12.If something ________ happens, it means that it doesn’t happen at any time.A.usually B.never C.sometimes13.— What’s your mother’s job, Amy?— She is an engineer. She is very busy, ________ she always goes home very late.A.but B.because C.so 14.—Xiaoming, can you help me? I can’t solve this ________ .—Of course. I will help you.A.pollution B.problem C.pattern15.— Do you usually ________ on foot, John?—No, I don’t. My home is a little far from my school, so I usually ride my bike there.A.go to bed B.go to school C.go to the club二、完形填空This morning, I book(预约)a taxi to the train station. When the car arrives, I just see theI say.The driver says, “Yes, ma’am.”I am 17 when I hear a woman’s voice! I 18 the driver. The driver is a young woman. In fact, I always 19 to different places by taxi, but I never see a woman driver here in India. In India, many people may think that is isn’t 20 for women to drive taxis, so I want to know more about my driver.“Hello, can I ask you some questions?” I ask.“Yes, ma’am.”“Are you well-educated(受过良好教育的)?”“Yes, ma’am. I got my MBA last year.”“Wow! So why did you take this 21 ?”She smiles and says, “ 22 is my interest, ma’am. No work is too small. I just want to 23 happiness from my work. I tried other work, but I like this one best. I think this is the most 24 work.”About 20 minutes later, I arrive. I say thank you to her. I really 25 the ride and learn something from the young woman.16.A.quickly B.quietly C.slowly 17.A.angry B.excited C.surprised 18.A.look for B.look at C.look after 19.A.stop B.travel C.take20.A.safe B.cool C.hard21.A.sound B.dream C.job22.A.Driving B.Singing C.Dancing 23.A.provide B.get C.protect 24.A.interesting B.difficult C.famous 25.A.need B.hate C.enjoy三、阅读单选Earth Day 2020We would like to invite you to be a very important part of our Earth Day Party. Activities:Music and Songs: Some classes in our school will sing songs about the Earth. These songs will help you know the importance of the Earth.Make a Recycled (循环利用的) Animal for Earth Day: You will learn about recycling things while making colourful animals. You can make a horse, a dog, a cat or a fish with your imagination (想象).Story Time: Our teacher will tell a story about the Earth. The story is For the Love of Our Earth. It is written by P.K. Hallinan.Date: Wednesday, April 22, 2020Time: 8:30 a.m.-3:30 p.mPlace: The Playground of Hope Middle SchoolEarth Day celebrates our beautiful Earth, the plants and the animals on the Earth! Thisholiday asks us to love the Earth and learn ways to protect it. Come and join us!26.What are the songs about?A.The party.B.A School.C.The Earth.D.The teacher. 27.What can you learn when making animals?A.Different kinds of animals.B.Different colours.C.Protecting animals.D.Recycling things.28.What is PK. Hallinan?A.A teacher.B.A student.C.A scientist.D.A writer. 29.The party will last for ________.A.six hours and a half B.seven hoursC.seven hours and a half D.eight hoursToday Mr. Green goes to school with his son Jim. He doesn’t often go to Jim’s school because he is very busy. But today, he goes there because it’s the school open day. Jim usually takes the school bus to school, but they go to school by car today. At the school gate, they see Mike and Julie. They’re the guidas(向导)today because their school is big and it’s difficult for some parents to find their children’s classrooms.The school open day is from 8:30 to 12:30. First, Mr. Green goes to a big meeting room. There’re many parents there. They talk with the teachers and drink tea. Then, the teachers show them around the school.At 10:30, they go to their children’s classrooms and watch a lesson. They play games with their children there. They have lunch in a large dining hall with their children. The lunch is delicious and the children are happy to have lunch with their parents at school!30.Why does Mr. Green go to school today?A.Because he is a teacher at this school.B.Because he isn’t busy today.C.Because he wants to have lunch with his son. D.Because it’s the school open day.31.How long does the school open day last?A.Three hours.B.Four hours.C.Five hours.D.All day. 32.Which do NOT the parents do on the school open day?A.Drink tea.B.Talk with teachers.C.Walk around the school.D.Give students a lesson.A.To tell us a busy day at school.B.To ask parents to talk with the teachers.C.To tell us the school open day.D.To let parents play with their children.I am Jenny. I was moved(感动)by a story last summer. When I went to a friend’s house as a guest. I found a sign on the door. It reads, “Before entering, please put down your trouble. When returning, bring back your happiness.”After entering the house, I saw the wife and the husband were both hamonious(和睦的), two children were friendly and polite and warmth filled the house.I asked them about the sign. The wife looked at the husband with a smile, “You tell the story.” The husband then looked back at the wife, “You say because it was your creativity(创意).”Finally, the wife said softly and slowly, “Once when I got home, I was surprised to see a sleepy and tired face in the elevator(电梯)mirror, with worried eyes. So I thought, when my children and husband faced this worried face, what would they feel? If I also faced such a face, what would I feel? Then I could imagine(想象)my children’s silence and my husaband’s indifferent(冷淡)at the dinner table. The next day, I wrote a sign and put it on the door. As a result, the sigh reminded(提醒)myself and the whole family. Something surprising happened in this way. Not only our family but also the guests to our house always become happy.”34.Who came up with the idea of putting the sign on the door?A.The wife.B.The husband.C.The whole family.D.The guest. 35.What does the words on the sign mean?A.A lot of people have worries at home.B.Everyone is tired when they come home because of their work.C.Everyone should have a happy face before entering the house.D.People may feel sleepy in the elevator.36.What does the underlined words “something surprising” mean?A.The children usually kept silent at home.B.The wife returned home with a worried face.C.The husband was indifferent at the dinner table.D.The guests to their house also become happy.A.Bring happiness back home.B.Learn to cook food for the family.C.Hang a wooden sign on the door.D.Help family members in trouble.There is a new school called Deep Green Bush School in New Zealand. The headmaster, called Moncarz, said to the teachers, “Please don’t tell kids it is time to do reading, writing or maths.”Headmaster Moncarz said, “If the weather is sunny, the students spend the day outdoors. They can play in the bush or on the grass, learn to fish or hunt, they can also pick apples and oranges and collect leaves and butterflies. At noon time, they cook lunch on an open fire. After school, they learn about the plants around their home or do something they like. There is no homework or classes. The students begin to learn more traditional(传统的)school skills, such as reading, writing and math only after they show an interest in them.”Well, is this school an “experiment” in education? Headmaster Moncarz answered no, “Most of the students are busy and unhappy in most traditional schools and their parents are worried about that. The idea comes from a lot of study. We just want the students to learn what they want to learn, do what they want to do.”38.The new school called Deep Green Bush School in the passage is in ________.A.Canada B.Australia C.New York D.New Zealand 39.When do the students do reading, writing or maths in Deep Green Bush School?A.After school.B.After they show an interest.C.At noon time.D.Before they show an interest. 40.When it is fine in the morning, the students in Deep Green Bush School can ________.A.sing B.shop C.pick fruit D.plant trees 41.What can we learn about Deep Green Bush School in New Zealand from the passage?A.It is a traditional school.B.It is an “experiment” in education.C.There’s no homework after school.D.There’re no teachers in the school.Everybody wants to learn English well but many find it not easy. I’m here to share my ideas about learning English with you.For some students, to remember new words is a big problem. In fact, words always come along with sentences(句子), so you can understand words in sentences. And you’d betterreview(复习)them now and then, or you will forget them easily.When reading, I like to write some special and important phrases down in the notebook so that I can remember them well and use them in my next reading. Studying by using is also a good way to learn English.After class, I go to the library sometimes. There are many books, newspapers and magazines in English. I usually read some of them for a long time there. It can help with my reading.At school, I go to the English club to talk with others in English every weekend. I am used to improving(提高)my oral English in this way.Do you know CCTV-10? I like to watch it whenever it’s possible. The programme is called Outlook is my favorite. In it, some teachers or scholars(学者)talk about helpful learning skills(技巧).We still have so much to learn about English. Never give up, and you’ll succeed one day. 42.What does Paragraph 2 mainly tell us?A.How to remember new words.B.How to understand words in sentences.C.How to use new words in reading.D.How to review words and use them well. 43.How many ways of learning English does the writer tell us?A.Three.B.Four.C.Five.D.Six. 44.What does the writer usually do to improve his reading?A.Remember new words in sentences.B.Take notes and read articles(文章)in English.C.Go to the English club.D.Watch English programmes of ten. 45.What does the writer mainly want to tell us?A.His school life B.An English clubC.A story about his school D.His ways of learning English四、根据句意填空50.—Which one do you like, tea coffee?—Tea, please.五、用所给单词的正确形式填空六、汉译英:整句56.杰克参加学校乐队的练习,他喜欢了解世界上不同的音乐风格。

广东省深圳市2020-2021学年宝安区七年级上学期期中联考数学试卷(PDF版 无答案)

广东省深圳市2020-2021学年宝安区七年级上学期期中联考数学试卷(PDF版 无答案)

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最新深圳市深圳中学数学七年级上册月考试卷及答案分析

最新深圳市深圳中学数学七年级上册月考试卷及答案分析

级上册月考试卷及答案分析第Ⅰ卷选择题(共30分)一、选择题(本大题共10个小题,每小题3分,共30分.在每个小题给出的四个选项中,只有一项符合题目要求,请选出并在答题卡上将该项涂黑)1、下列结论中正确的是()A.0既是正数,又是负数B.O是最小的正数C.0是最大的负数 D.0既不是正数,也不是负数2. 12点15分,钟表上时针与分针所成的夹角的度数为A.B.C.D.3.已知,,则与的大小关系是A.B.C.D.无法确定4.如果一个角的余角是50°,则这个角的补角的度数是A.130°B.140°C.40°D.150°5.给出四个数:-1,1/3,0.5, 1/7,其中为无理数的是()A.-1 B.1/3 C.0.5 D.1/76.把弯曲的道路改直,能够缩短行程,其道理用数学知识解释应是……………( ) A.垂线段最短B.两点确定一条直线C.线段可以大小比较D.两点之间,线段最短7.下面几何体的主视图是( )正面 A B C D8.已知m≥2,n≥2,且m、n均为正整数,如果将m n进行如图所示的“分解”,那么下列四个叙述中正确的有………………………………()①在25的“分解”中,最大的数是11.②在43的“分解”中,最小的数是13.③若m3的“分解”中最小的数是23,则m=5.④若3n的“分解”中最小的数是79,则n=5.A.1个B.2个C.3个D.4个9............a......b..2............A.2B.b.a C.a.b D.b.a+2 10.....a.b.c...............a..b.c.........( ) A..b.c..aB..b..a.c C..a.c..b D..a..b.c第Ⅱ卷非选择题(共90分)二、填空题(本大题共5个小题,每小题3分,共15分)11、如果节约16度电记作+16度,那么浪费5度电记作度;12.计算:17254'︒⨯= .13.多项式-3xy44+3x+26的最高次项系数是__________.14.当x= _________ 时,代数式4x+2与3x﹣9的值互为相反数.15.某超市推出如下优惠方案:(1)一次性购物不超过100元不享受优惠;(2)一次性购物超过100元但不超过300元一律九折;(3)一次性购物超过300元一律八折。

2020-2021深圳市外国语学校数学七年级上入学试卷及答案分析

2020-2021深圳市外国语学校数学七年级上入学试卷及答案分析

级上入学试卷及答案分析第Ⅰ卷选择题(共30分)一、选择题(本大题共10个小题,每小题3分,共30分.在每个小题给出的四个选项中,只有一项符合题目要求,请选出并在答题卡上将该项涂黑)1.几何体的下列性质:①侧面是平行四边形;②底面形状相同;③底面平行;④棱长相等.其中棱柱具有的性质有A.1个B.2个C.3个D.4个2.下列各式计算正确的是 --------------------------------------------------------------------------- ()A.=-6;B.(-3)2=-9;C.-3 2=-9;D. -(-3)2=93.如果用+3表示运入仓库的大米吨数,那么运出5 t大米可表示为()A.-5 t B.+5 t C.-3 t D.+3 t4、右图是“东方”超市中“飘柔”洗发水的价格标签,一服务员不小心将墨水滴在标签上,使得原价看不清楚,请帮忙算一算,该洗发水的原价是:()A. 22元B. 23元C. 24元D. 26元5.一个数和它的倒数相等,则这个数是( )A.1B.﹣1C.±1D.±1和06.一个两位数,十位上的数字是x,个位上的数字比十位上的数字的2倍少3,这个两位数可以表示为…………………………………………………………………………()A.x(2x-3) B.x(2x+3) C.12x+3 D.12x-37.如图,C,D是线段AB上两点,若CB=4cm,DB=7cm,且D是AC的中点,则AC的长等于……………………………………………………………()A.3 cm B.6 cm C.11 cm D.14 cmA BCD(第7题)8.下列合并同类项中,正确的是( )A.2a+3b=5ab B.5b2﹣2b2=3C.3ab﹣3ba=0D.7a+a=7a29、已知线段AB=6,在直线AB上取一点C,使BC=2,则线段AC的的()A.2B.4 C.8 D.8或410.下面每个表格中的四个数都是按相同规律填写的:根据此规律确定x的值为( )A.135B.170C.209D.252第Ⅱ卷非选择题(共90分)二、填空题(本大题共5个小题,每小题3分,共15分)11、如果节约16度电记作+16度,那么浪费5度电记作度;12.光的传播速度大约是300 000 000米/秒,用科学记数法可表示为米/秒.x-3的几何意义是数轴上表示数x的点与表示数3的点之间的距离,则式13.我们知道:式子||子||x+1的最小值为.x-2+||14.如图,延长线段AB到C,使BC=4,若AB=8,则线段AC的长是BC的_________倍.15.下边横排有12个方格,每个方格都有一个数字,若任何相邻三个数字的和都是20,则x =.5 A B C D E F x G H I 10三、解答题(本大题共7个小题,共75分.解答应写出文字说明、证明过程或演算步骤)16.计算(1))(-12)-5+(-14)-(-39);(2)(3)17.(本题共8分,每小题4分)(1)已知:A=2m2+n2+2m,B=m2-n2-m,求A-2B的值.(2)先化简,再求值:5a2-[3a-2(2a-1)+4a2],其中a=-.18.已知代数式9-6y-4y2=7,求2y2+3y+7的值.19.李师傅打算把一个长、宽、高分别为50cm,8cm,20cm的长方体铁块锻造成一个立方体铁块,问锻造成的立方体铁块的棱长是多少cm?20.如图,纸上有五个边长为1的小正方形组成的图形纸,我们可以把它剪开拼成一个正方形。

广东省深圳市宝安区 新安中学(集团)外国语学校2019-2020学年七年级下册第一次联考测试数学

广东省深圳市宝安区 新安中学(集团)外国语学校2019-2020学年七年级下册第一次联考测试数学


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广东省深圳市宝安区2021-2021学年七年级第二学期第一次月考数学试卷

广东省深圳市宝安区2021-2021学年七年级第二学期第一次月考数学试卷

. 2021—2021学年第二学期月考考试卷七年级数学 〔时间:90分钟 总分值:100分〕 一、选择题:本大题共10小题,每题3分,共30分;在每题给出的四个选项中,只有一项为哪一项符合题目要求的. 1、以下各题中计算错误的选项是〔 〕 ()()323321818A m n m n ⎡⎤--=-⎢⎥⎣⎦、 322398()()B m n mn m n --=-、 ()322366()C m n m n ⎡⎤--=-⎣⎦、 232399()()D m n mn m n --=、 2、化简x(y-x)-y(x-y)得〔 〕 A 、x 2-y 2 B 、y 2-x 2 C 、2xy D 、-2xy ()()2000199919992 1.513⎛⎫⨯⨯- ⎪⎝⎭的结果是〔 〕 A .23 B .-23 C .32 D .-32 4.1622++ax x 是一个完全平方式,那么a 的值为〔 〕 A.4 ——8 5.02267,56,43⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛⎪⎭⎫ ⎝⎛-三个数中,最大的是〔 〕 A.243-⎪⎭⎫ ⎝⎛ B.256⎪⎭⎫ ⎝⎛ C.067⎪⎭⎫ ⎝⎛ 6.化简〔a+b+c 〕2-〔a -b+c 〕2的结果为〔 〕 A.4ab+4bc B.4ac C.2ac D.4ab -4bc 7.3181=a ,4127=b ,619=c ,那么a 、b 、c 的大小关系是〔 〕 A .a >b >c B .a >c >b C .a <b <c D .b >c >a 8.假设142-=y x ,1327+=x y ,那么y x -等于〔 〕 A .-5 9.边长为a 的正方形,边长减少b 以后所得较小正方形的面积比原来正方形的面积减少了〔 〕 A .2b B .2b +2ab C .2ab D .b (2a —b ) 10.多项式251244522+++-x y xy x 的最小值为〔 〕 A .4 B .5 C .16 D .25 二、填空题:本大题共6小题,每题3分,共18分,把答案填写在题中横线上. 11. 1022223x x y π--+-是_____次_____项式,常数项是_____,最高次项是_____.12.〔1〕912327( ) a b -= 〔2〕23294,272,3____m n m n --===则13. 〔1〕(21)(12)_____x x ---= 〔2〕22(2)(24)_____a b a ab b -+++= 14.2249x mxy y -+是关于,x y 的完全平方式,那么m = ; 15.假设m 2+n 2-6n +4m +13=0,m 2-n 2= ;16、如果3=x 时,代数式13++qx px 的值为2021,那么当3-=x 时,代数式13++qx px 的值是三、计算题:本大题共5小题,每题4分,共20分,解容许写出必要的计算过程. 17.2202211(2)()()[(2)]22----+---+--;18.32236222()()()()xx x x x ÷+÷-÷-19.22222(32)(32)(94)x y x y x y -++20.(322)(322)m n m n -+++21.221(2)(2)(2)(2)()()n n x y y x x y x y x y x y --÷-+---+--+四、综合题:本大题共5小题,共32分,解容许写出必要的计算过程.22.〔5分〕2514x x -=,求()()()212111x x x ---++的值23.〔6分〕简便计算:〔1〕.1234612344123452⨯-42×2.24.〔5分〕20052009a x =+,20052010b x =+,20052011c x =+,求代数式ca bc ab c b a ---++222的值;25.〔6分〕假设4m 2+n 2-6n +4m +10=0,求n m - 的值;26.〔8分〕假设2228()(3)03x px x x q ++-+=的积中不含2x 与3x 项, 〔1〕求p 、q 的值;〔2〕求代数式23120102012(2)(3)pq pq p q --++的值;答案:1-5.CBBCA; 6-10.AABDC; 11.23,323,;x y π-- 12.〔1〕343a b -〔2〕29; 13. 〔1〕214x -〔2〕338b a -;14. 12+-; 15.-5;16、-2006;17.5316;18.2; 19. 442(8116)x y -;20. 2291244m m n ++-;21.2252y x xxy --- ; 25.-8; 26.173,,(2)21539p q ==-;。

广东省深圳市外国语学校七年级上期中考试数学试题(无答案)

广东省深圳市外国语学校七年级上期中考试数学试题(无答案)

广东省深圳市外国语学校七年级上期中考试数学试题(无答案)数学一、选择题(每题3分,共36分)1.a 与21-互为倒数,那么a 是( ) A.2 B.-2 C.21- D.21 2.2021年5月5日,奥运火炬手携带着意味〝战争、友谊、提高〞的奥运圣火火种,分开海拔5200米的〝珠峰大本营〞,向山顶攀爬。

他们在海拔每上升100米,气温就下降0.6℃的低平和缺氧的状况下,于5月8日9时17分,成功登上海拔8844.43米的地球最高点。

而此时〝珠峰大本营〞的温度为-4°C,峰顶的温度为(结果保管整数)( )A.-26℃B.-22℃C.-18℃D.-22℃3.设a 是最小的自然数,b 是最大的负整数,c 是相对值最小的有理,那么c b a +-的值为( )A.-1B.0C.1D.24.以下说法正确的选项是( )A.0既不是正数,也不是正数,所以0不是有理数B.在-3与-1之间仅有一个有理数C.一个正数的倒数一定还是正数D.一个数的相对值越大,表示它的点在数轴上越靠右5.以下单项式中,与23.0ab 是同类项的是( )A.b a 22B.22b aC.a b 241- D.ab 3D. 3ab 6.以下有理数()()014.35126,,,,------,其中正数的个数有( )A.1个B.2个C.3个D.4个7.以下说法中错误的选项是( 〕A.单项式xyz 5.0的次数为3B.单项式32vt -的系数是-2 C.15与31-同类项 D.ab a 211--是二次三项式 8.把()()()()7539+----+-写成省略括号的代数和方式是( )A.7539--+-B.7539-+--C.7539+++-D.7539-++-9.假定532-+x x 的值为7,那么2932-+x x 的值为( )A.0B.24C.34D.4410.如图,从边长为()4+a cm 的正方形纸片中剪去一个边长为()1+a cm 的正方形(a >0),剩余局部沿虚线又剪拼成一个长方形(不堆叠无缝隙),那么长方形的面积为( )A.()22cm 52a a + B.()2cm 156+a C.()2cm 96+a D.()2cm 153+a11.一个两位数的十位数字为a ,个位数字比十位数字的2倍少1,假定把这个两位数十位上的数字与个位上的数字交流位置组成一个新两位数,那么原两位数与新两位数的差为( )A.a 99-B.1111-aC.99-aD.1133-a12.假定用A 、B 、C 区分表示有理数c b a 、、,O 为原点如下图.化简a c a b c a ---+-的结果为( )A.c b a -+2B.c a b 23+-C.c b a 2-+D.a b -二、填空题(每题3分,共18分)13.假设〝盈利10%〞记作+10%,那么〝盈余20%〞记作___________.14.P 是数轴上的一个点,它到原点的距离是4个单位,那么P 点表示的数是_______. 15.3=a ,且0=+a a ,那么=+-12a a __________.16.找出以下各图形中数的规律,依此规律,a 的值为_________.17.观察表格中按规律陈列的两行数据,假定用y x ,表示表格中间一列的两个数,那么y x ,满足的数量关系是________________. 18.25==b a ,.假定a >0,b <0,12=-c ,那么c b ab c -+2的值为_____________. 三、解答题(共46分)19.(16分)计算与化简:(1)()()48.25243÷--⨯-+ (2)()3661397125-⨯⎪⎭⎫ ⎝⎛-+- (3)()⎥⎦⎤⎢⎣⎡--+⎪⎭⎫ ⎝⎛⨯---42.06.0511222 (4)()b b a b a 33453922122-⎪⎭⎫ ⎝⎛---- 20.(6分)先化简,再求值:事先32-==x m ,,求()⎪⎭⎫ ⎝⎛-----+-mx mx mx mx 311333122. 21.(6分)初一某班抽查了10名同窗的期中数学考试效果,以80分为规范,超出80分的都记为正数,缺乏80分的局部记为正数,结果如下:+8,-3,+15,-7,-5,+9,-8,+1,0,+10.(1)这10名同窗中最高分是___________,最低分是___________;(2)求这10名同窗的平均效果。

广东省深圳市宝安区外国语学校(集团)2024—2025学年上学期七年级10月月考数学试卷

广东省深圳市宝安区外国语学校(集团)2024—2025学年上学期七年级10月月考数学试卷

广东省深圳市宝安区外国语学校(集团)2024—2025学年上学期七年级10月月考数学试卷一、单选题1.12024-的相反数是( ) A .12024 B .12024- C .2024 D .2024-2.在2-, 2.4+,13-,0.72,124-,0, 1.8-中.非负数共有( ) A .3个 B .2个 C .1个 D .0个3.下列数轴表示正确的是( )A .B .C .D . 4.若用一个平面截一个正方体得到的截面是三角形,则该三角形一定是( ) A .锐角三角形 B .直角三角形 C .钝角三角形 D .无法确定 5.下表是国外城市与北京的时差(带正号的数表示同一时刻比北京时间早的时数).那么与北京时间最接近的城市是( )A .伦敦B .墨尔本C .东京D .巴黎 6.如图①所示的是一个正方体的表面展开图,将对应的正方体从如图②所示的位置依次翻过第1格、第2格,到第3格时正方体朝上的一面上的字是( )A .亚B .欢C .迎D .您7.如图,点A ,B 在数轴上对应的有理数分别为m ,n ,若点A 向右移动x 个单位长度后到达B 点,则x 的值为( )A .mB .m n -C .n m -D .n8.现有一列数1a ,2a ,3a ,L ,48a ,49a ,50a ,其中32020a =,72018a =-,471a =-,并且满足任意相邻三个数的和为同一个常数,则123484950a a a a a a ++++++L 的值为( )A .2035-B .2035C .2003-D .2003二、填空题9.李白出生于公元701年,我们记作+701,那么秦始皇出生于公元前256年,可记作. 10.比较大小:32-213-.(用“>”“=”或“<”填空). 11.钟表上的时针转动一周形成一个圆面,这说明了 .12.若有理数a ,b 满足340a b ++-=,则ab =.13.有理数a 、b 、c 在数轴上的对应点如图所示,化简:b c b a c -++-=.三、解答题14.计算: (1)()()67 3.215⎛⎫----+- ⎪⎝⎭; (2)()32.1 6.57⎛⎫-⨯⨯- ⎪⎝⎭; (3)()1113612366⎛⎫--+⨯- ⎪⎝⎭; (4)11992412⎛⎫-⨯ ⎪⎝⎭; (5)()124215⎛⎫÷-÷- ⎪⎝⎭;(6)()4211.5554⎛⎫-⨯÷-⨯ ⎪⎝⎭. 15.画出数抽,在数轴上表示下列各数,并用“>”把它们连接起来.11420(1)|3|323⎛⎫----+-- ⎪⎝⎭,,,,, 16.如图,是由一些大小相同的小正方体组合成的简单几何体,请分别画出从正面、左面和上面观察该几何体看到的形状图.17.已知a ,b 互为相反数,c ,d 互为倒数,m 的绝对值等于4,p 是数轴上原点表示的数.(1)分别直接写出a b +,cd ,m ,p 的值; (2)a b p cd m cd+-++的值是多少? 18.深圳地铁11号线()Shenzhen Metro Line 11是深圳市境内第6条建成运营的地铁线路,于2016年6月28日开通运营,深圳地铁11号线起于岗厦北站,途经福田区、南山区、宝安区,贯穿大空港地区、城市商务区,前海、后海片区,止于碧头站,其中的8个站点如图所示.小红从福永站开始乘坐地铁,在图中8个地铁站点做值勤志愿服务,到A 站下车时,本次志愿者活动结束,约定向碧头站方向为正,当天的乘车记录如下(单位:站):5,2,3,2,3,1,4,2++-+--+-.(1)请你通过计算说明A 站是哪一站?(2)已知相邻两站之间的平均距离为1.5千米,求小红在志愿者服务期间乘坐地铁行进的路程是多少千米?19.下表记录的是流花河今年某一周内的水位变化情况,上周末(星期六)的水位已达到警戒水位33米.(正号表示水位比前一天上升,负号表示水位比前一天下降)(1)本周哪一天河流的水位最高?哪一天河流的水位最低?它们位于警戒水位之上还是之下?(2)与上周末相比,本周末河流的水位是上升了还是下降了?(3)以警戒水位作为零点,用折线统计图表示本周的水位情况.A B C为数轴上的三点,若点C到点A的距20.利用数轴可以将“数与形”完美地结合,已知,,,A B的“优点”.离是点C到点B的距离的2倍,我们就称点C是()A B C三点所表示的数分别为1-,2,1,点C到点A的距离AC=(1)例如,如图1,数轴上点,,,A B的______,点C到点B的距离是BC=______,因为AC是BC的两倍,所以称点C是()“优点”.,E F(2)如图2,,,E F P为数轴上三点,点E所表示的数为0,点F所表示的数为3,当点P是()的“优点”时,求此时点P表示的数是多少?(3)如图3,,G H为数轴上两点,点G所表示的数为30-,点H所表示的数为90.现有一电子蚂蚁Q从点G出发,以2个单位每秒的速度向右运动,到达点H停止.当运动时间t为何G H Q三个点中,恰有一个点为其余两点的“优点”?值时,,,。

2020-2021学年广东省深圳市宝安区新安中学(集团)外国语学校七上第二次月考数学试卷

2020-2021学年广东省深圳市宝安区新安中学(集团)外国语学校七上第二次月考数学试卷

2020-2021学年广东省深圳市宝安区新安中学(集团)外国语学校七上月考数学试卷一、选择题(共12小题;共60分)1. 在,,,这四个数中,负数有A. 个B. 个C. 个D. 个2. 的倒数是A. B. C. D.3. 下列有理数大小关系判断正确的是A. B.C. D.4. 的值是A. B. C. D.5. 把写成省略括号后的算式为A. B.C. D.6. 如果有理数,满足条件,那么的值是A. 正数B. 负数C. 非负数D. 非正数7. 下列说法正确的是A. 所有的整数都是正数B. 不是正数的数一定是负数C. 是最小的有理数D. 整数和分数统称有理数8. 计算时,可以使运算简便的是A. 乘法交换律B. 乘法分配律C. 加法结合律D. 乘法结合律9. 数,,,中最大的是A. B. C. D.10. 计算,结果正确的是A. B. C. D.11. 近年来,国家重视精准扶贫,收效显著,据统计,约有人脱贫.将用科学记数法表示应为A. B. C. D.12. 若规定“”是一种数学运算符号,且,,,,则的值为A. B. C. D.二、填空题(共4小题;共20分)13. 我市某天的最高气温是,最低气温是,则这天的日温差是.14. .绝对值与倒数均等于它本身的数是.15. 已知,则.16. 的计算结果,用以为底的幂的形式表示是.三、解答题(共7小题;共91分)17. 计算:(1).(2).18. 有筐白菜,以每筐为标准,超过或不足的千克数分别用正、负数来表示,记录如下:(1)筐白菜中,最重的一筐比最轻的一筐重多少千克?(2)与标准质量相比,筐白菜总计超过或不足多少千克?(3)若白菜每千克售价元,则出售这筐白菜可卖多少元?19. 把下列各数分别填在相应的大括号内:,,,,,,,,.(1)整数有:;(2)分数有:;(3)正整数有:;(4)负整数有:;(5)正分数有:;(6)负分数有:;20. 已知:数轴上,两点表示的有理数分别为,,且.(1)求的值.(2)数轴上的点分别与,两点的距离的和为,求点在数轴上表示的数的值.21. 计算:(1);(2);(3);(4).22. 已知,.(1)写出和的值;(2)若,求的值.23. 阅读的计算方法.解:设原式的商为,因为,所以,所以商.请按以上方法计算.答案第一部分1. A 【解析】;;;,负数有个.2. B3. C4. D5. D6. A7. D 【解析】A、负整数和就不是正数,显然A错误;B、不是正数,有可能是零,所以B错误;C、负有理数比零小,错误;D、正确.8. B9. A10. C11. A12. C第二部分13.14.【解析】绝对值与倒数均等于它本身的数是.故答案为:.15.16.第三部分17. (1).(2).18. (1)最重的一筐比最轻的一筐重.(2).答:与标准质量相比,筐白菜总计不足.(3)(元).答:出售这筐白菜可卖元.19. (1),,,,(2),,(3),(4),(5),(6)20. (1),,,解得,,;(2),,数轴上,两点表示的有理数分别为,,数轴上的点与,两点的距离的和为,点可能在点的左侧或点可能在点的右侧,当点在点的左侧时,,得,当点在点的右侧时,,得,即点在数轴上表示的数的值是或.21. (1).(2).(3)(4)22. (1),,,.(2)当时,,或,,此时.23. 设原式的商为,,,商.。

2020-2021深圳市外国语学校数学七年级第四月考试题及答案分析下载

2020-2021深圳市外国语学校数学七年级第四月考试题及答案分析下载

2020-2021深圳市外国语学校数学七年级第四月考试题及答案分析下载第Ⅰ卷选择题(共30分)一、选择题(本大题共10个小题,每小题3分,共30分.在每个小题给出的四个选项中,只有一项符合题目要求,请选出并在答题卡上将该项涂黑)1.-2的相反数是()A.2 B.-2 C.0 D.42.据报道,我省西环高铁预计2015年底建成通车,计划总投资27100000000元,数据27100000000用科学记数法表示为( )A.271×108B.2.71×109C.2.71×1010 D.2.71×10113.在数轴上与-3的距离等于4的点表示的数是().A.1B.-7C.1或-7D.无数个4、右图是“东方”超市中“飘柔”洗发水的价格标签,一服务员不小心将墨水滴在标签上,使得原价看不清楚,请帮忙算一算,该洗发水的原价是:()A. 22元B. 23元C. 24元D. 26元5. 一种面粉的质量标识为“25±0.25千克”,则下列面粉中合格的是( )A.24. 70千克B.25.30千克C.24.80千克D.25.51千克6.一个两位数的两个数字之和为7,则符合条件的两位数的个数是……………( ) A.8 B.7 C.6 D.57.如图,在下列四个几何体中,它的三视图(主视图、左视图、俯视图)不完全相同的是…………………………………………………………………………………( )①正方体②圆柱③圆锥④球A .①②B .②③C .①④D . ②④8.下列合并同类项中,正确的是( )A .2a+3b=5abB .5b 2﹣2b 2=3C .3ab ﹣3ba=0D .7a+a=7a 29.下列说法正确的有( )①两点确定一条直线;②两点之间线段最短;③∠α+∠β=90°,则∠α和∠β互余;④一条直线把一个角分成两个相等的角,这条直线叫做角的平分线.A .1个B .2个C .3个D .4个10.下列各组算式中,其值最小的是( )A .﹣(﹣3﹣2)2B . (﹣3)×(﹣2)C . (﹣3)2×(﹣2)D . (﹣3)2÷(﹣2)第Ⅱ卷 非选择题(共90分)二、填空题(本大题共5个小题,每小题3分,共15分)11.(4分)2.5的相反数是 ,的倒数是 .12.计算:17254'︒⨯= .13.绝对值大于1而小于4的整数的和是 ;积为14.若单项式3a 5b m +1与-2a n b 2是同类项,那么m +n = . 15.一个机器人从数轴原点出发,沿数轴正方向,以每前进3步后退2步,不断往返的程序运动.设该机器人每秒钟前进或后退1步,并且每步的距离为1个单位长,x n 表示第n 秒时机器人在数轴上的位置所对应的数.则下列结论:(1)x 3=3;(2)x 8=4;(3)x 105<x 104 ;(4)x 2013<x 2014 中,正确结论的个数是_______________.三、解答题 (本大题共7个小题,共75分.解答应写出文字说明、证明过程或演算步骤)16.计算:(1) (34 + 56-712)÷ 124 (2) -14-|-5| + 8× (-12) 217. 解方程:(每小题4分,共8分)(1) 8x =12(x -2); (2)2x +13-5x -16=118.已知||a -1+||ab -2=0,求代数式1ab +1(a +1)(b +1)+1(a +2)(b +2)+…+1(a +2014)(b +2014)的值.19.李师傅打算把一个长、宽、高分别为50cm ,8cm ,20cm 的长方体铁块锻造成一个立方体铁块,问锻造成的立方体铁块的棱长是多少cm ?20.用长为10m 的铝合金做成如图的长方形窗框,设窗框横档的长为m ,中间一条直档与横档长度相等.(1)用含的代数式表示这个窗户的面积(中间的横档与直档所占的面积忽略不计); (2)当横档长取1.4m 时,求窗户的面积.21 .如图,将连续的奇数1、3、5、7 …… ,排列成如下的数表,用十字框框出5个数。

深圳市宝安区2020—2021学年七年级上第二次月考数学试卷含解析

深圳市宝安区2020—2021学年七年级上第二次月考数学试卷含解析

深圳市宝安区2020—2021学年七年级上第二次月考数学试卷含解析数学试卷一、填空题(每小题只有一个正确的选项,每小题3分,共36分)1.下面两个数互为相反数的是()A.﹣(+7)与+(﹣7)B.﹣0.5与﹣(+0.5)C.﹣1.25与D.+(﹣0.01)与﹣(﹣)2.科学记数法a×10n中a的取值范畴为()A.0<|a|<10 B.1<|a|<10 C.1≤|a|<9 D.1≤|a|<103.如图的几何体是下面()平面图形绕轴旋转一周得到的.A.B.C.D.4.一个正方体的侧面展开图有几个全等的正方形()A.2个 B.3个 C.4个 D.6个5.下列合并同类项中,正确的是()A.3x+3y=6xy B.2a2+3a3=5a3C.3mn﹣2mn=mn D.7x﹣5x=26.a、b在数轴上的位置如图,则所表示的数是()A.a是正数,b是负数B.a是负数,b是正数C.a、b差不多上正数D.a、b差不多上负数7.已知(﹣2x2+3)3=a0+a1(x﹣1)+a2(x﹣1)2+a3(x﹣1)3+…+a6(x﹣1)6,则a0+a6=()A.﹣5 B.﹣6 C.﹣7 D.﹣88.假如减数为负数,则()A.差比被减数小B.差比被减数大C.差为正数D.差为负数9.下列各组有理数的大小比较中,不正确的是()A.﹣(﹣8)>﹣8 B.C.D.﹣(﹣1.414)>0 10.下列说法中不正确的是()A.若a为任一有理数,则a的倒数是B.若|a|=|b|,则a=±bC.x2=(﹣2)2,则x=±2D.x2+1一定是正数11.4x3﹣(﹣2x)3+(﹣9x3)的值是()A.﹣3x3B.x3C.3x3D.5x312.若代数式2x﹣y的值是5,则代数式2y﹣4x+5的值为()A.﹣15 B.﹣5 C.5 D.15二、选择题(每小题3分,共12分)13.某条河流的最高水位是58.4米,戒备水位是55.1米,把它的戒备水位作为0点,则最高水位用有理数表示为米.14.单项式﹣的次数是,系数是.15.若﹣x2y m+1与﹣x n y2是同类项,那么m=,n.16.(2+1)(22+1)(24+1)…(22048+1)的个位数是.三、解答题(共52分)17.(16分)一项工程,甲单独做5天能够完成全工程;假如乙,丙两队合作12天能够完成全工程;假如三队合作,多少天能够完成全工程?18.(8分)合并同类项:5y﹣2x2y﹣3y+3x2y.19.(5分)求解:x﹣2(x2﹣y2)+(2x﹣2y2),其中x=﹣3,y=﹣2.20.(6分)依照立体图从上面看到的形状图(如图所示),画出它从正面和左面看到的形状图(图中数字代表该位置的小正方体的个数).21.(5分)若m>0,n<0,|n|>|m|,用“<”号连接m,n,0,|n|,﹣m,请结合数轴解答.22.(6分)某市第5路公交车从起点到终点共有8个站,一辆公交车由起点开往终点,在起点站始发时上了部分乘客,从第二站开始下车、上车的乘客数如下表:二三四五六七八站次人数下车(人)24375816上车(人)7864350(1)求起点站上车人数;(2)若公交车收费标准为上车每人2元,运算此趟公交车从起点到终点的总收入;(3)公交车在哪两个站之间运行时车内乘客最多?是几人?23.(6分)如图,长方形ABCD的长为a,宽为b,分别以A,B为圆心,以AD,BC为半径作两个圆.(1)用代数式表示阴影部分的周长和面积;(2)当a=8,b=3时,求阴影部分的周长和面积.2021-2020学年广东省深圳市宝安区七年级(上)第二次月考数学试卷参考答案与试题解析一、填空题(每小题只有一个正确的选项,每小题3分,共36分)1.(3分)下面两个数互为相反数的是()A.﹣(+7)与+(﹣7)B.﹣0.5与﹣(+0.5)C.﹣1.25与D.+(﹣0.01)与﹣(﹣)【分析】本题需依照相反数的概念,对每一项进行分析,即可求出正确答案.【解答】解:A、∵﹣(+7)=﹣7,+(﹣7)=﹣7∴﹣(+7)和+(﹣7)不互为相反数,故本选项错误;B、∵﹣(+0.5)=﹣0.5,∴﹣0.5和﹣(+0.5)不互为相反数,故本选项错误;C、∵=0.8,∴﹣1.25和不互为相反数,故本选项错误;D、∵+(﹣0.01)=﹣0.01,﹣(﹣)=0.01,∴+(﹣0.01)与﹣(﹣)互为相反数,故本选项错误正确.故选:D.【点评】本题考查了相反数的定义,是基础题,熟记概念是解题的关键.2.(3分)科学记数法a×10n中a的取值范畴为()A.0<|a|<10 B.1<|a|<10 C.1≤|a|<9 D.1≤|a|<10【分析】科学记数法确实是将一个数字表示成(a×10的n次幂的形式),其中1≤|a|<10,n表示整数.【解答】解:科学记数法a×10n中a的取值范畴为1≤|a|<10.故选D.【点评】本题考查科学记数法的定义,是需要熟记的内容.3.(3分)如图的几何体是下面()平面图形绕轴旋转一周得到的.A.B.C.D.【分析】依照面动成体的原理即可解,一个三角形绕直角边旋转一周能够得到一个圆锥.【解答】解:圆锥的轴截面是直角三角形,因而圆锥能够认为直角三角形以一条直角边所在的直线为轴旋转一周得到.故选B.【点评】本题要紧考查空间观念,难度不大,学生应注意培养空间想象能力.4.(3分)一个正方体的侧面展开图有几个全等的正方形()A.2个 B.3个 C.4个 D.6个【分析】可把一个正方体展开,观看侧面全等的正方形的个数即可.【解答】解:因为一个正方体的侧面展开会产生4个完全相等的正方形,因此有4个全等的正方形.故选C.【点评】本题考查的是全等形的识别,属于较容易的基础题.5.(3分)下列合并同类项中,正确的是()A.3x+3y=6xy B.2a2+3a3=5a3C.3mn﹣2mn=mn D.7x﹣5x=2【分析】依照同类项的定义先判定是否为同类项,假如是依照合并同类项的法则进行运算.【解答】解:A.因为3x与3y不是同类项,故3x+3y不能合并,故选项错误;B.因为2a2与3a3不是同类项,故它们不能合并,故选项错误;C.因为3mn﹣2mn=mn,故选项正确;D.因为7x﹣5x=2x,故选项错误.故选C.【点评】本题考查同类项的定义和合并同类项的法则.6.(3分)a、b在数轴上的位置如图,则所表示的数是()A.a是正数,b是负数B.a是负数,b是正数C.a、b差不多上正数D.a、b差不多上负数【分析】依照数轴的特点进行解答即可.【解答】解:∵由图可知,a在原点的左侧,b在原点的右侧,∴a为负数,b为正数.故选B.【点评】本题考查的是数轴,熟知数轴的特点是解答此题的关键.7.(3分)已知(﹣2x2+3)3=a0+a1(x﹣1)+a2(x﹣1)2+a3(x﹣1)3+…+a6(x ﹣1)6,则a0+a6=()A.﹣5 B.﹣6 C.﹣7 D.﹣8【分析】把x=1和x=0代入代数式分析解答即可.【解答】解:把x=1代入(﹣2x2+3)3=a0+a1(x﹣1)+a2(x﹣1)2+a3(x﹣1)3+…+a6(x﹣1)6,可得:1=a0,把x=0代入(﹣2x2+3)3=a0+a1(x﹣1)+a2(x﹣1)2+a3(x﹣1)3+…+a6(x﹣1)6,可得:27=a0﹣a1+a2﹣a3+…+a6,把x=2代入(﹣2x2+3)3=a0+a1(x﹣1)+a2(x﹣1)2+a3(x﹣1)3+…+a6(x﹣1)6,可得:﹣27=a0+a1+a2+a3+…+a6,27﹣27=2a0+2a2+2a4+2a6;27+27=﹣2a1﹣2a3﹣2a5可得:a0+a6=﹣7;故选C【点评】本题要紧考查代数式求值,利用赋值法是解决本题的关键.8.(3分)假如减数为负数,则()A.差比被减数小B.差比被减数大C.差为正数D.差为负数【分析】依照有理数的减法运算,减去一个数等于加上那个数的相反数解答.【解答】解:∵减数为负数,∴相当于加上一个正数,∴差比被减数大.故选B.【点评】本题考查了有理数的减法,熟记减去一个数等于加上那个数的相反数是解题的关键.9.(3分)下列各组有理数的大小比较中,不正确的是()A.﹣(﹣8)>﹣8 B.C.D.﹣(﹣1.414)>0【分析】第一化简有理数,然后依照有理数大小比较法则即可得出答案.【解答】解:A、﹣(﹣8)=8,8>﹣8,正确;B、﹣(﹣)=4.5,4.5=4.5,不正确;C、+(﹣1)=﹣1,﹣1正确;D、﹣(﹣1.414)=1.414,1.414>0,正确;不正确的是B;故选B.【点评】此题考查了有理数的大小比较,把握正数大于负数,假如两数差不多上正数,则绝对值大的大,绝对值小的小,假如两数差不多上负数,则绝对值大的数反而小.10.(3分)下列说法中不正确的是()A.若a为任一有理数,则a的倒数是B.若|a|=|b|,则a=±bC.x2=(﹣2)2,则x=±2D.x2+1一定是正数【分析】各项利用有理数的乘方,绝对值的代数意义,倒数的定义,以及非负数性质判定即可.【解答】解:A、若a为不为0的有理数,则a的倒数是,不符合题意;B、若|a|=|b|,则a=±b,符合题意;C、x2=(﹣2)2=4,则x=±2,符合题意;D、x2+1一定是正数,符合题意,故选A【点评】此题考查了有理数的乘方,绝对值,倒数,以及非负数的性质:偶次幂,熟练把握各自的性质是解本题的关键.11.(3分)4x3﹣(﹣2x)3+(﹣9x3)的值是()A.﹣3x3B.x3C.3x3D.5x3【分析】先去括号,再合并同类项即可.【解答】解:原式=4x3+8x3﹣9x3=(4+8﹣9)x3=3x3.故选C.【点评】本题考查的是整式的加减,熟知整式的加减实质上确实是合并同类项是解答此题的关键.12.(3分)若代数式2x﹣y的值是5,则代数式2y﹣4x+5的值为()A.﹣15 B.﹣5 C.5 D.15【分析】依照相反数的意义,可得(y﹣2x)的值,依照代数式求值,可得答案.【解答】解:由2x﹣y的值是5,得y﹣2x=﹣5.2y﹣4x+5=2(y﹣2x)+5=2×(﹣5)+5=﹣5,故选:B.【点评】本题考查了代数式求值,利用(y﹣2x)整体代入是解题关键.二、选择题(每小题3分,共12分)13.(3分)某条河流的最高水位是58.4米,戒备水位是55.1米,把它的戒备水位作为0点,则最高水位用有理数表示为+3.3米.【分析】戒备水位作为0点,以上就正数.最高水位是58.4米应表示为:58.4米﹣55.1米=+3.3米.【解答】解:由题意得,最高水位是58.4米应表示为:58.4米﹣55.1米=+3.3米.【点评】考查正负数在生活中的应用.14.(3分)单项式﹣的次数是4,系数是﹣.【分析】利用单项式的次数与系数的定义求解即可.【解答】解:单项式﹣的次数是4,系数是﹣.故答案为:4,﹣.【点评】本题要紧考查了单项式,解题的关键是熟记单项式的次数与系数的定义.15.(3分)若﹣x2y m+1与﹣x n y2是同类项,那么m=1,n2.【分析】依照同类项的概念求解.【解答】解:∵﹣x2y m+1与﹣x n y2是同类项,∴n=2,m+1=2,∴m=1,n=2.故答案为:1,2.【点评】本题考查了同类项的知识,解答本题的关键是把握同类项定义中的两个“相同”:相同字母的指数相同.16.(3分)(2+1)(22+1)(24+1)…(22048+1)的个位数是5.【分析】原式中的1变形为2﹣1,反复利用平方差公式运算即可得到结果.【解答】解:(2+1)(22+1)(24+1)…(22048+1)=(2﹣1)(2+1)(22+1)(24+1)…(22048+1)=(22﹣1)(22+1)(24+1)(28+1)…(22048+1)=(24﹣1)(24+1)(28+1)…(22048+1)=24096﹣1,∵21=2,22=4,23=8,24=16,25=32,…,∴个位上数字以2,4,8,6为循环节循环,∵4096÷4=1024,∴24096个位上数字为6,即原式个位上数字为5.故答案为:5【点评】此题考查了平方差公式,熟练把握平方差公式是解本题的关键.三、解答题(共52分)17.(16分)一项工程,甲单独做5天能够完成全工程;假如乙,丙两队合作12天能够完成全工程;假如三队合作,多少天能够完成全工程?【分析】把这项工程的工作总量看作单位“1”,甲的工作效率为,乙、丙两队的工作效率和为,进一步求得三个队的工作效率和,利用工作总量÷工作效率=工作时刻列式解答即可.【解答】解:1÷(+)=1÷=(天)答:假如三队合作,天能够完成全工程.【点评】此题考查有理数的混合运算的实际运用,把握工作效率、工作总量、工作时刻三者之间的关系是解决问题的关键.18.(8分)合并同类项:5y﹣2x2y﹣3y+3x2y.【分析】原式合并同类项即可得到结果.【解答】解:原式=2y+x2y.【点评】此题考查了合并同类项,熟练把握合并同类项法则是解本题的关键.19.(5分)求解:x﹣2(x2﹣y2)+(2x﹣2y2),其中x=﹣3,y=﹣2.【分析】依照去括号、合并同类项,可化简整式,依照代数式求值的方法,可得答案.【解答】解:原式=x﹣2x2+y2+2x﹣2y2=﹣2x2+3x﹣y2.当x=﹣3,y=﹣2时,原式=﹣2×(﹣3)2+3×(﹣3)﹣(﹣2)2=﹣2×9+3×(﹣3)﹣4=18﹣9﹣4=5.【点评】本题考查了整式的化简求值,去括号:括号前是负号去掉括号要变号,括号前是正号去掉括号不变号.20.(6分)依照立体图从上面看到的形状图(如图所示),画出它从正面和左面看到的形状图(图中数字代表该位置的小正方体的个数).【分析】由已知条件可知,从正面看有2列,每列小正方数形数目分别为3,4;从左面看有2列,每列小正方形数目分别为2,4.据此可画出图形.【解答】解:如图所示:【点评】此题考查几何体的三视图画法.由几何体的俯视图及小正方形内的数字,可知主视图的列数与俯视数的列数相同,且每列小正方形数目为俯视图中该列小正方形数字中的最大数字.左视图的列数与俯视图的行数相同,且每列小正方形数目为俯视图中相应行中正方形数字中的最大数字.21.(5分)若m>0,n<0,|n|>|m|,用“<”号连接m,n,0,|n|,﹣m,请结合数轴解答.【分析】第一依照在数轴上表示数的方法,利用数轴标出m,n的大致位置,再标出﹣m,﹣n的大致位置;然后依照当数轴方向朝右时,右边的数总比左边的数大,把这些数由小到大用“<”号连接起来即可.【解答】解:,n<﹣m<0<m<﹣n.【点评】此题要紧考查了有理数大小比较的方法,以及数轴的特点:一样来说,当数轴方向朝右时,右边的数总比左边的数大,要熟练把握.22.(6分)某市第5路公交车从起点到终点共有8个站,一辆公交车由起点开往终点,在起点站始发时上了部分乘客,从第二站开始下车、上车的乘客数如下表:站次二三四五六七八人数下车(人)24375816上车(人)7864350(1)求起点站上车人数;(2)若公交车收费标准为上车每人2元,运算此趟公交车从起点到终点的总收入;(3)公交车在哪两个站之间运行时车内乘客最多?是几人?【分析】(1)依照下车的总人数减去上车的总人数得到起点站上车的人数即可;(2)依照表格运算得出此趟公交车从起点到终点的总收入即可;(3)依照表格得出二站到三站上车的乘客最多,是8人.【解答】解:(1)依照题意得:(2+4+3+7+5+8+16)﹣(7+8+6+4+3+5)=45﹣33=12(人),则起始站上车12人;(2)依照题意得:依照题意得:2(12+7+8+6+4+3+5)=90(元),则此趟公交车从起点到终点的总收入为90元;(3)依照表格得:七站到八站上车的乘客最多,是24人.【点评】此题考查了正数与负数,弄清题意是解本题的关键.23.(6分)如图,长方形ABCD的长为a,宽为b,分别以A,B为圆心,以AD,BC为半径作两个圆.(1)用代数式表示阴影部分的周长和面积;(2)当a=8,b=3时,求阴影部分的周长和面积.【分析】(1)依照图形可得阴影部分的周长是×2πb+a+(a﹣2b);阴影部分的面积为长方形的面积减去两个圆的面积(半圆的面积)即可;(2)将a、b的值代入(1)中所列代数式求值即可.【解答】解:(1)阴影部分的周长L=×2πb+a+(a﹣2b)=πb+2a﹣2b;阴影部分的面积S=ab﹣πb2;(2)当a=8,b=3时,L=3π+16﹣6=10+3π,S=8×3﹣π×9=24﹣π.【点评】此题考查的是列代数式和代数式求值,用到的知识点是半圆的周长和面积的运算方法.。

精品解析:广东省深圳市外国语学校2020-2021学年七年级上学期期末数学试题(解析版)

精品解析:广东省深圳市外国语学校2020-2021学年七年级上学期期末数学试题(解析版)

深圳外国语学校 2020-2021 第一学期期末考试初一数学试卷一、选择题(共 10 题,每题 3 分,共 30 分)1. -2的倒数是( )A. -2B. 2C. 12D. 12- 【答案】D【解析】【分析】根据倒数的定义即可求解.【详解】-2的倒数是12-故选D .【点睛】此题主要考查倒数的求解,解题的关键是熟知有理数的性质.2. 下列各式运算正确的是( )A. 2x 35x +=B. 23a 5a 8a +=C. 223a b 2a b 1-=D. 22ab b a 0-= 【答案】D【解析】【分析】利用并同类项的法则判定即可.【详解】A 、2x+3不是同类项不能加减,故本选项错误,B 、3a+5a =8a ,故本选项错误,C 、3a 2b ﹣2a 2b =a 2b ,故本选项错误,D 、ab 2﹣b 2a =0,故本选项正确,故选D .【点睛】本题主要考查了合并同类项,解题的关键是熟记合并同类项的法则.3. 下列调查方式,你认为最合适的是( )A. 日光灯管厂要检测一批灯管的使用寿命,采用全面调查方式B. 旅客上飞机前的安检,采用抽样调查方式C. 了解郑州市居民日平均用水量,采用全面调查方式D. 了解郑州市每天的平均用电量,采用抽样调查方式【答案】D【解析】【分析】由普查得到的调查结果比较准确,但所费人力、物力和时间较多,而调查得到的调查结果比较近似,据此作答;【详解】日光灯管厂要检测一批灯管的使用寿命,应用抽样调查,故A错误;旅客上飞机前的安检,采用普查方式,故B错误;了解郑州市居民日平均用水量,应用抽样调查,故C错误;了解郑州市每天的平均用电量,采用抽样调查方式,故D正确;故答案是D.【点睛】本题主要考查了全面调查与抽样调查,准确分析是解题的关键.4. 如图,是一个几何体的表面展开图,则该几何体中写“英”的面相对面上的字是( )A. 战B. 疫C. 情D. 颂【答案】B【解析】【分析】正方体的表面展开图,相对的面之间一定相隔一个正方形,根据这一特点作答.【详解】解:正方体的表面展开图,相对的面之间一定相隔一个正方形,“战”与“情”是相对面,“疫”与“英”是相对面,“颂”与“雄”是相对面.故选:B.【点睛】本题主要考查了正方体相对两个面上的文字,注意正方体的空间图形,从相对面入手分析是解题的关键.5. 桂林是世界著名的风景旅游城市和历史文化名城,地处南岭山系西南部,广西东北部,行政区域总面积27 809平方公里.将27 809用科学记数法表示应为()A. 0.278 09×105B. 27.809×103C. 2.780 9×103D. 2.780 9×104【答案】D【解析】【分析】科学记数法的表示形式为a×10n的形式,其中1≤|a|<10,n为整数.确定n的值时,要看把原数变成a 时,小数点移动了多少位,n的绝对值与小数点移动的位数相同.当原数绝对值大于10时,n是正数;当原数的绝对值小于1时,n是负数.10,故选D.【详解】27 809=2.780 9×4【点睛】本题考查了科学记数法的表示方法.科学记数法的表示形式为a×10n的形式,其中1≤|a|<10,n为整数,表示时关键要正确确定a的值以及n的值6. 下列说法正确的个数是()(1)连接两点之间的线段叫两点间的距离;(2)两点之间,线段最短;(3)若AB=2CB,则点C是AB的中点;(4)角的大小与角的两边的长短无关.A. 1个B. 2个C. 3个D. 4个【答案】B【解析】①因为“连接两点的线段的长度叫做两点间的距离”,所以(1)中说法错误;②因为“两点之间,线段最短”,所以(2)中说法正确;③因为:当点C不在AB上时,即使AB=2CB时,点C也不是AB中点,所以(3)中说法错误;④因为“角的大小与角的边长无关”,所以(4)中说法正确;即上述说法中,正确的说法有2个,是(2)、(4).故选B.7. 如图,甲、乙两人同时从A 地出发,甲沿北偏东50 方向步行前进,乙沿图示方向步行前进.当甲到达B 地,乙到达C 地时,甲与乙前进方向的夹角∠BAC 为100︒,则此时乙位于A地的( )A. 南偏东30︒B. 南偏东50︒C. 北偏西30︒D. 北偏西50︒ 【答案】A【解析】【分析】直接根据题意得出各角度数,进而结合方向角表示方法得出答案.【详解】解:如图所示:由题意得:∠1=50︒,∠BAC =100︒∴∠2=180°-∠1-∠BAC=180°-50︒-100︒=30︒故乙位于A地南偏东30︒.故选:A.【点睛】此题主要考查了方向角,正确掌握方向角的表示方法是解题关键.8. 定义a*b=ab+a+b,若3*x=27,则x的值是()A. 3B. 4C. 6D. 9【答案】C【解析】【分析】根据运算规则转化为一元一次方程,然后解即可.【详解】解:根据运算规则可知:3*x=27可化为3x+3+x=27,移项可得:4x=24,即x=6.故选C .【点睛】本题考查解一元一次方程的解法;解一元一次方程常见的思路有通分,移项,左右同乘除等. 9. 钟表在8点30 分时,时钟上的时针与分针之间的夹角为( )A. 60︒B. 70︒C. 75︒D. 85︒【答案】C【解析】【分析】根据钟表上12个数字,每相邻两个数字之间的夹角为30°计算得到答案.【详解】解:8点30分,时针和分针中间相差2.5个大格.∵钟表12个数字,每相邻两个数字之间的夹角为30°,∴8点30分分针与时针的夹角是2.5×30°=75°.故选C .【点睛】本题考查了钟面角,用到的知识点为:钟表上12个数字,每相邻两个数字之间的夹角为30°. 10. 小明在解方程513m x -=(x 为未知数)时,误将x -看作x +,得方程的解为2x =-,原方程的解为( )A. 0x =B. 1x =C. 2x =D. 3x = 【答案】C【解析】【分析】把x =−2代入方程513m x +=,求出m ,得出方程为15−x =13,求出方程的解即可.【详解】解:把x =−2代入方程513m x +=得:5m−2=13,解得m =3,即原方程为15−x =13,解得x =2.故选:C .【点睛】本题考查了一元一次方程的解和解一元一次方程,根据方程的解的定义能求出m 的值是解此题的关键.二、填空题(共 5 小题,每题 3 分,共 15 分)11. 数轴上,点B 在点A 的右边,已知点A 表示的数是﹣2,且AB =5.那么点B 表示的数是_____.【答案】3.【解析】【分析】根据数轴表示数的意义,在点A 的右边,到点A 距离为5的点所表示的数为3.【详解】解:﹣2+5=3,故答案为:3.【点睛】本题考查数轴表示数的意义和方法,在数轴表示的数右边总比左边的大.12. 某种商品每件的进价为120元,标价为180元.为了拓展销路,商店准备打折销售.若使利润率为20%,则商店应打________折.【答案】八【解析】【分析】打折销售后要保证打折后利率为20%,因而可以得到不等关系为:利润率=20%,设可以打x 折,根据不等关系列出不等式求解即可.【详解】解:设应打x 折,则根据题意得:(180×x×10%-120)÷120=20%, 解得:x=8.故商店应打八折.故答案为:八.【点睛】本题考查一元一次方程的实际应用,解题关键是读懂题意,找到符合题意的等量关系式,同时要注意掌握利润率的计算方法.13. 如图,线段4AB cm =,延长线段AB 到C ,使1BC cm =,再反向延长AB 到D ,使3,AD cm E =是AD 中点,F 是CD 的中点.则EF 的长度为._____cm .【答案】2.5【解析】【分析】根据条件可以求出线段DC的长,因为E是AD中点,F是CD的中点,所以可求出线段DE,DF的长,而EF=DF-DE,所以EF=2.5cm.【详解】解:由AB=4cm,BC=1cm ,AD=3cm得DC=4+1+3=8cm因为E是AD的中点,所以DE=12AD=1.5cm因为F是CD的中点,所以DF=12CD=4cm所以EF=DF-DE=2.5cm 故答案为:2.5 【点睛】本题考查的是线段长度的相关计算,根据对应图形并进行线段的和、差计算是解题的关键.14. 将相同的长方形卡片按如图方式摆放在一个直角上,每个长方形卡片长为2,宽为1,依此类推,摆放2019个时,实线部分长为______.【答案】5048【解析】【分析】确定实线部分长的变化规律,根据此规律即可确定摆放第2019个时的实线部分长.【详解】解:摆放1个时实线部分长为3;摆放2个时实线部分长为3231215+=⨯+⨯=;摆放3个时实线部分长为3233228++=⨯+=;摆放4个时实线部分长为3232322210+++=⨯+⨯=;摆放5个时实线部分长为32323332213++++=⨯+⨯=;摆放6个时实线部分长323232332315+++++=⨯+⨯=;……摆放2019个时实线部分长为31010210095048⨯+⨯=.故答案为:5048.【点睛】本题主要考查了图形的变化规律,灵活根据已有条件确定变化规律是解题的关键.15. 已知∠AOB=20°,∠AOC=4∠AOB,OD平分∠AOB,OM平分∠AOC,则∠MOD的度数是_____________________度【答案】30或50【解析】【分析】本题分两种情况讨论:(1)当C、B两点位于OA边的同一侧时;(2)当当C、B两点位于OA边的两侧时.根据两种情况,同时考虑角平分线的定义,得出结论即可.【详解】分为两种情况:如下图,当∠AOB在∠AOC内部时,∵∠AOB=20°,∠AOC=4∠AOB,∴∠AOC=80°,∵OD平分∠AOB,OM平分∠AOC,∴∠AOD=∠BOD=12∠AOB=10°,∠AOM=∠COM=12∠AOC=40°,∴∠DOM=∠AOM-∠AOD=40°-10°=30°. 如下图,当∠AOB在∠AOC外部时,∵∠AOB=20°,∠AOC=4∠AOB ,∴∠AOC=80°,∵OD 平分∠AOB ,OM 平分∠AOC ,∴∠AOD=∠BOD=12∠AOB=10°,∠AOM=∠COM=12∠AOC=40°,∴∠DOM=∠AOM+∠AOD=40°+10°=50°. 【点睛】本题考查角平分线的应用等相关知识,解题时不能只考虑一种情况,要分当∠AOB 在∠AOC 内部和当∠AOB 在∠AOC 外部两种情况进行讨论.三、解答题(共 7 小题,共 55 分)16. 计算:()()()220210324125π-+⨯---+- 【答案】-7【解析】【分析】先依据零指数幂的性质、有理数的乘方、绝对值法则计算,最后算加减即可点.【详解】解:原式=4-4-8+1=-7.【点睛】本题主要考查的是零指数幂的性质、有理数的乘方、绝对值法则计算熟练掌握相关知识是解题的关键.17. 解方程:5717323x x ++-= 【答案】3113x = 【解析】【分析】这是一个带分母方程,所以要先去分母,再去括号,最后移项,化系数为1,从而得到方程的解.【详解】解:去分母得:()()35721718x x +-+=152123418x x +--=,1331x =, 解得:3113x =. 【点睛】本题考查的是解一元一次方程,需要熟练掌握解一元一次方程的步骤.18. 先化简,再求值:()()2222232233y x x xy x y-+--+的值,其中x 1,y 2==-.【答案】-6xy ,12.【解析】【分析】去括号、合并同类项化简后代入求值即可.【详解】3y 2−x 2+2(2x 2−3xy )−3(x 2+y 2)=3y 2−x 2+4x 2−6xy−3x 2−3y 2=−6xy ,当x =1,y =−2时,原式=−6×1×(−2)=12.【点睛】本题考查整式的加减,去括号、合并同类项是整式加减的基本方法.19. 某市教育局为了了解初一学生第一学期参加社会实践活动的情况,随机抽查了本市部分初一学生第一学期参加社会实践活动的天数,并将得到的数据绘制成了下面两幅不完整的统计图.请根据图中提供的信息,回答下列问题:(1)扇形统计图中 a 的值为 ,该扇形圆心角的度数为 ;(2)补全条形统计图;(3)如果该市共有初一学生 20000 人,请你估计“活动时间不少于 5 天”的大约有多少人?【答案】(1)25%;90°;(2)见解析;(3)15000人【解析】【分析】(1)用1减去其他天数所占的百分比即可得到a 的值,用360°乘以它所占的百分比,即可求出该扇形所对圆心角的度数;(2)先求出参加社会实践活动的总人数,再乘以参加社会实践活动为6天的所占的百分比,求出参加社会实践活动为6天的人数,从而补全统计图;(3)用总人数乘以活动时间不少于5天的人数所占的百分比即可求出答案.【详解】解:(1)扇形统计图中a=1-30%-15%-10%-20%=25%,该扇形所对圆心角的度数为360°×25%=90°;故答案为:25%,90°;(2)参加社会实践活动的总人数是:2020010%(人),则参加社会实践活动为6天的人数是:200×25%=50(人),补图如下:(3)该市初一学生第一学期社会实践活动时间不少于5天的人数约是:20000×(30%+25%+20%)=15000(人).【点睛】本题考查了条形统计图、扇形统计图等知识,读懂统计图,从不同的统计图中得到必要的信息是解决问题的关键.条形统计图能清楚地表示出每个项目的数据;扇形统计图直接反映部分占总体的百分比大小.20. 一个旅游团共26人去参观一个景点,已知成人票每张120元,儿童票每张80元,经预算,共需要门票钱2640元.(1)求这个旅游团成人和儿童的数量各是多少人?(2)到了售票窗口得知,购买两张成人票将会赠送一张儿童票,请计算共需门票钱多少元?【答案】(1)这个旅游团成人的数量是14人,儿童的数量是12人;(2)共需门票2080元.【解析】【分析】(1)设旅游团成人的数量是x人,则儿童的数量是(26-x)人,根据成人票的费用加上儿童票的费用等于2640,解方程即可;(2)用2640减去优惠金额即可.【详解】(1)设旅游团成人的数量是x人,则儿童的数量是(26–x)人,根据题意得120x+80(26–x)=2640,解得x=14,26–x=26–14=12.答:这个旅游团成人的数量是14人,儿童的数量是12人;(2)2640–14÷2×80=2080(元).答:共需门票2080元.【点睛】本题考查了一元一次方程在实际问题中的应用,正确分析题意得出方程,是解题的关键.21. 如图,点A,B,C 在数轴上表示的数分别是-3,3 和1.动点P ,Q 同时出发,动点P 从点A 出发,以每秒6个单位的速度沿A →B向终点B匀速运动;动点Q 从点C 出发,以每秒1个单位的速度沿C →B 向终点B 匀速运动,当P、Q都到达终点后停止运动.设点P 的运动时间为t(s) .(1)当点P 到达点B 时,点Q 所表示的数是;(2)当t= 0.5时,线段PQ 的长为;(3)在整个运动过程中,当P ,Q 两点到点C 的距离相等时,求t 的值.【答案】(1)2;(2)1.5;(3)47或45或2.【解析】【分析】(1)根据两点之间的距离公式,时间=路程÷速度,路程=速度×时间,列式计算即可求解;(2)求出t=0.5时,P、Q点所表示的数,再根据两点间的距离公式可求线段PQ的长;(3)分3种情况讨论可求t的值.【详解】解:(1)∵点P 到达点B所需时间=[3-(-3)]÷6=1,∴点Q 所表示的数=1×1+1=2.故答案是:2;(2)当t= 0.5时,点Q 所表示的数是(1×0.5+1)=1.5,点P 所表示的数是(-3+6×0.5)=0.∴线段PQ的长是1.5;故答案是:1.5;(3)①相遇前,依题意有1-(-3+6t)=t,解得t=47;②当P追上Q时,依题意有(6-1)t=1-(-3),解得t=45;③当P到达B停止运动后, Q到达B时,此时P ,Q 两点到点C 的距离相等,依题意有t=2.综上所述,t的值为47或45或2.【点睛】考查了一元一次方程的应用及数轴上两点的距离,掌握路程问题等量关系和数轴知识是解题的关键.22. 如图1,射线OC 在∠AOB 的内部,图中共有3 个角:∠AOB 、∠AOC 和∠BOC ,若其中有一个角的度数是另一个角度数的两倍,则称射线OC 是∠AOB 的奇妙线.(1)一个角的角平分线这个角的奇妙线.(填是或不是)(2)如图2,若∠MPN = 60︒,射线PQ 绕点P 从PN 位置开始,以每秒10︒的速度逆时针旋转,当∠QPN 首次等于180︒时停止旋转,设旋转的时间为t(s) .①当t 为何值时,射线PM 是∠QPN 的奇妙线?②若射线PM 同时绕点P 以每秒6︒的速度逆时针旋转,并与PQ 同时停止旋转.请求出当射线PQ 是∠MPN 的奇妙线时t 的值.【答案】(1)是;(2)①9或12或18;②52或307或203【解析】【分析】(1)根据奇妙线定义即可求解;(2)①分3种情况,∠QPN=2∠MPN;∠MPN=2∠QPM;∠QPM =2∠MPN.列出方程求解即可;②分3种情况,∠MPN=2∠QPN;∠MPQ=2∠QPN;∠QPN =2∠MPQ.列出方程求解即可.【详解】(1)设∠α被角平分线分成的两个角为∠1和∠2,则有∠α=2∠1,∴一个角的平分线是这个角的“奇妙线”;故答案是:是;(2)①由题意可知射线PM 在∠QPN的内部,∴∠QPN=(10t)︒,∠QPM=(10t-60)︒,(a)当∠QPN=2∠MPN时,10t=2×60,解得t=12;(b)当∠MPN=2∠QPM时,60=2×(10t-60),解得t=9;(c)当∠QPM =2∠MPN时,(10t-60)=2×60,解得t=18.故当t为9或12或18时,射线PM是∠QPN的“奇妙线”;②由题意可知射线PQ 在∠MPN的内部,∴∠QPN=(10t)︒,∠MPN=(60+6t)︒,∠QPM=∠MPN-∠QPN=(60-4t)︒,(a)当∠MPN=2∠QPN时,60+6t=2×10t,解得t=307;(b)当∠MPQ=2∠QPN时,60-4t=2×10t,解得t=52;(c)当∠QPN =2∠MPQ时,10t=2×(60-4t),解得t=203.故当射线PQ是∠MPN的奇妙线时t的值为52或307或203.【点睛】本题考查了角之间的关系及一元一次方程的应用,奇妙线定义,学生的阅读理解能力及知识的迁移能力.理解“奇妙线”的定义是解题的关键.。

2020-2021学年深圳宝安新安中学初中部高三英语月考试卷及答案

2020-2021学年深圳宝安新安中学初中部高三英语月考试卷及答案

2020-2021学年深圳宝安新安中学初中部高三英语月考试卷及答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项ATheatre reflects the values of the civilization out of which it grows. The following are the types of theatre performances an ancient Roman might have witnessed then.Fescennine VerseFescennine Verse was a pioneer of Roman comedy. Ironic and improvisational(即兴的), it was used mainly at festivals or weddings, and as invective. With early native Italian funny dialogues in Latin verse, it was thought to have combined with a tradition of performances by masked dancers and musicians from Etruria.Fabula AtellanaFabula Atellana relied on common characters, masks, direct humor, and simple plots. They were performed by actors improvising. Fabula Atellana came from the Oscan city of Atella. There were 4 main types of characters: the braggart, the greedy blockhead, the clever hunchback and the stupid old man, like modern Punch and Judy shows.Fabula TogataNamed for the clothing symbolic of the Roman people Fabula Togata had various subtypes. One was the Fabula Tabernaria, named for the tavern(酒馆)where the comedy’s preferred characters, lowlifes, might be found. One describing more middle-class types, and continuing the Roman clothing theme, was the Fabula Trabeata.Fabula PraetextaFabula Praetexta is the name for Roman tragedies on Roman themes, Roman history or current politics. Fabula Praetexta was less popular than tragedies on Greek themes. During the Golden Age of drama in the Middle Republic, there were four great Roman writers of tragedy, Naevius, Ennius, Pacuvius, and Accius. Of their surviving tragedies, 90 titles remain.All the performances above began as a translation of Greek forms, even to the extent of their being performed in Greek costume.1.Where might an ancient Roman witness Fescennine Verse?A.At a party.B.At a funeral.C.At a wedding.D.At a concert.2.Which type of performance describes the middle-class life?A.Fabula Atellana.B.Fabula Tabernaria.C.Fabula Trabeata.D.Fabula Praetexta.3.What do the listed types of performances have in common?A.They copy Latin dramas.B.They take on Greek forms.C.They reflect Roman themes.D.They refer to Italian stories.BIn 2015, a man named Nigel Richards memorized 386, 000 words in the entireFrench Scrabble Dictionaryin just nine weeks. However, he does not speak French. Richards’ impressive feat is a useful example to show how artificial intelligence works — real AI. Both of Richard and AI take in massive amounts of data to achieve goals with unlimited memory and superman accuracy in a certain field.The potential applications for AI are extremely exciting. Because AI canoutperformhumans at routine tasks — provided the task is in one field with a lot of data — it is technically capable of replacing hundreds of millions of white and blue collar jobs in the next 15 years or so.But not every job will be replaced by AI. In fact, four types of jobs are not at risk at all. First, there are creative jobs. AI needs to be given a goal to optimize. It cannot invent, like scientists, novelists and artists can. Second, the complex, strategic jobs — executives, diplomats, economists — go well beyond the AI limitation of single-field and Big Data. Then there are the as-yet-unknown jobs that will be created by AI.Are you worried that these three types of jobs won’t employ as many people as AI will replace? Not to worry, as the fourth type is much larger: jobs where emotions are needed, such as teachers, nannies and doctors. These jobs require compassion, trust and sympathy — which AI does not have. And even if AI tried to fake it, nobody would want a robot telling them they have cancer, or a robot to babysit their children.So there will still be jobs in the age of AI. The key then must be retraining the workforce so people can do them. This must be the responsibility not just of the government, which can provide funds, but also of corporations and those who benefit most.4. What is the main purpose of paragraph 1?A. To introduce the topic.B. To mention Nigel’s feat.C. To stress the importance of good memory.D. To suggest humans go beyond AI in memory.5. Which of the following best explains “outperform” underlined in paragraph 2?A. Be superior toB. Be equal toC. Be similar toD. Be related to6. Which of the following jobs is the most likely to be replaced?A. The writer.B. The shop assistant.C. The babysitter.D. The psychologist.7. What does the text suggest people do about job replacement of AI?A. Limit the application of AI to a certain degree.B. Get more support from the government.C. Apply for the donation from companies.D. Upgrade themselves all the time.CSam, I say to myself as I start across the bridge, you must stop these thoughts and start thinking about what to do now that you have lost your falcon, Frightful.Life, my friend Ban do once said, is meeting problems and solving them whether you are an amoeba or a space traveller. I have a problem. I have to provide my younger sister Alice and myself with meat. Fish, nuts, and vegetables are good and necessary, but they don't provide enough fuel for the hard physical work we do. Although we have venison now, I can't always count on getting it. So far this year, our venison has been only road kill from in front of Mrs Strawberry's farm.I decide to take the longest way home, down the flood plain of the West Branch of Delaware to Spillkill, my own name for a fast stream that cascades down the south face of the mountain range I'm on. I need time to think. Perhaps Alice and I should be like the early Eskimos. We should walk, camp and hunt, and when the seasons change, walk on to new food sources. But I love my tree and my mountaintop.Another solution would be to become farmers, like the people of the Iroquois Confederacy who once lived here. They settled in villages and planted corm and squash, bush beans and berries. We already grow groundnuts in the damp soil and squash in the poor land. But the Iroquois also hunted game. I can't do that anymore.I'm back where I started from.Slowly I climb the Spillkill. As I hop from rock to rock beneath shady basswoods and hemlocks, I hear the cry of the red-tailed hawk who nests on the mountain crest. I am reminded of Frightful and my heart aches. I can almost hear her call my name, Cree, Cree, Cree, Car-ree.Maybe I can get her back if I beg the man who is in charge of the peregrines at the university. “But it's the law,” he would say. I could write to the president of the United States and ask him to make an exception of Alice and me. That won't work. The president swore to uphold the Constitution and laws of the United States when he took office.I climb on. I must stop thinking about the impossible and solve the problem of what to do now. I must find a new way to provide for us. Frightful is going to be in good hands at the university, and she will have young.I smile at the thought of little Frightfuls and lift my reluctant feet.When I am far above the river, I take off my clothes and moccasins and bathe in a deep, clear pool until I am refreshed and thinking more clearly. Climbing up the bank, I dress and sit down. I breathe deeply of the mountain air and try to solve my problem more realistically.8. What does this excerpt main describe?A. Delicate mental activities.B. Unique story environment.C. Everchanging story events.D. Complicated character relationship.9. What is Sam's first worry?A. How to get back quicklyB. How to get enough venison.C. How to ensure the safety of Frightful.D. How to provide meat for Alice and himself.10. What do we know about Frightful?A. He left Sam and Alice due to lack of food.B. He helped Sam hunt before being taken away.C. He is living with the red-tailed hawk happily.D. He has given birth to babies in the university.11. Which of the following can best describe Sam?A. Humorous.B. Aggressive.C. Responsible.D. Unrealistic.DThe English language is changing, and you are responsible! Whether we consider changes in grammar, spelling, pronunciation, or the very vocabulary of the language, you have played your part and continue to do so.When we first learned basic grammar and spelling, perhaps in elementary school, we might have gotten the impression that these things were sacred. The rules that apply to such things might have been presented as unchanging and unchangeable. While this way might be helpful for teaching children, it is far from accurate.The English language, like many others, is a living, growing, ever-evolving thing. Like it or not, you are involved in this change. These changes take many forms. Grammar and spelling have changed greatly over theyears and centuries, with the spelling differences in different countries today a reflection of this. While the language of a thousand years ago might be called English, most of us would hardly recognize it today as the same language.The first involves changes in the pronunciation of words. Many are familiar with the differences between the British and American ways of pronouncing certain words. In addition to these differences, the pronunciation of many words has changed over the years because of how you have decided to pronounce them. For example, consider the word "err." The traditional pronunciation of this word rhymes with the word "her." Older dictionaries show this to be the primary or only pronunciation. However, in recent years, more and more people have been pronouncing it so that it sounds like "air." Another change in the language involves the addition and removal of words. The makers of dictionaries decide which words deserve to be officially adopted as part of the English language. Through the centuries, many words have come from other languages. In fact, English has probably done this more than any other language in the world, which is why spelling and pronunciation rules for English have so many exceptions.Of course, many slang words have been just short-lived fashions that have died out quickly. Others, though, have been adopted by mainstream society and become respectable, as have many technical terms. So then remember, the next time you repeat the newest expression to hit the street, or make up your own words, you may be contributing to the future of the English language.12. When we begin to learn English, we think _________.A. it is interesting to pick up a new languageB. English rules are wrongly presented in factC. grammar and spelling rules are unchangeableD. only adults have the ability to affect a language13. From paragraph 3 we can know that_________.A. we can change the English languageB. many languages are changing over yearsC. English has changed little in the past 1,000 yearsD. there were main changes in grammar and pronunciation14. Why is the pronunciation of words changing?A. people speak in different waysB. people have adopted foreign wordsC. it has been affected by American EnglishD. makers of dictionaries often change them15. Which of the following is the best title for the passage?A. Foreign words involved in English.B. The British speaks differently from Americans.C. English language is changing over years.D. You can change the English language.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

2020-2021广东深圳市宝安中学数学七年级上册第四月考试卷(含答案)

2020-2021广东深圳市宝安中学数学七年级上册第四月考试卷(含答案)

2020-2021广东深圳市宝安中学数学七年级上册第四月考试卷(含答案)第Ⅰ卷 选择题(共30分)一、选择题(本大题共10个小题,每小题3分,共30分.在每个小题给出的四个选项中,只有一项符合题目要求,请选出并在答题卡上将该项涂黑)1.实效m ,n 在数轴上的对应点如图所示,则下列各式子正确的是( )A .B .C .D .2.如果水位下降3米记作-3米,那么水位上升4米,记作( )A 、1米.B 、7米.C 、4米.D 、-7米. 3.如果用+3表示运入仓库的大米吨数,那么运出5 t 大米可表示为( )A .-5 tB .+5 tC .-3 tD .+3 t4、下列计算中,错误的是( )。

A 、B 、C 、D 、5.下列计算正确的是…………………………………………………………………( )A .-3(a +b )=-3a +3bB .2(x +12y )=2x +12y C .x 3+2x 5=3x 8 D .-x 3+3x 3=2x 3 6.关于x 的方程2x +a -8=0的解是x =2,则a 的值是………………………………………( )A .2B .3C .4D .57.如图,在下列四个几何体中,它的三视图(主视图、左视图、俯视图)不完全相同的是 …………………………………………………………………………………( )①正方体 ②圆柱 ③圆锥④球A.①②B.②③ C.①④D.②④8.一个长方形的周长为20,其中它的长为a,那么该长方形的面积是…………()A.20a B.a(20-a) C.10a D.a(10-a)9.已知x=1是关于x的方程2-ax=x+a的解,则a的值是()A.B.C.D.110.有一个班去划船,计划租若干条船,这时班长说,若再增加一条船,则每条船坐6人,若减少一条船,则每条船坐9人,这个班共有( )人.A.32 B.36 C.40 D.48第Ⅱ卷非选择题(共90分)二、填空题(本大题共5个小题,每小题3分,共15分)11、x^2=64,则x= ____________。

2020-2021学年深圳宝安新安中学初中部高三英语月考试卷及参考答案

2020-2021学年深圳宝安新安中学初中部高三英语月考试卷及参考答案

2020-2021学年深圳宝安新安中学初中部高三英语月考试卷及参考答案第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AWhen the sun shines brightly, it provides a great chance to get outdoor things done. Like making hay! At least, that is what farmers from the past would say. ―Make hay while the sun shines.This idiom is very old, dating back to Medieval times. Rain would often ruin the process of making hay. So, farmers had no choice but to make hay when the sun was shining.Today, we all use this expression, not just farmers. When conditions are perfect to get something done, we can say, ―It’s a good idea to make hay while the sun shines.In other words, you are taking advantage of a good situation or of good conditions. You are making the most of your opportunities. These all mean ―making hay while the sun shines.And sometimes we use this expression to mean we beat someone to the punch, or we got ahead of someone else. And other times you make hay while the sun shines to make good use of the chance to do something while it lasts. You are being opportunistic – taking advantage of a good opportunity. For example, my friend Ozzy was sick for a week and could not go to work. So, his co-worker Sarah -- who doesn’t like him -- took advantage of his illness and stole his project! Talk about making hay while the sun shines.Sometimes when you make hay while the sun shines you are staying ahead of a problem – like in this example:Hey, do you want to go hiking with me and my friends this weekend? The weather is going to be beautiful! I wish I could. But I have to finish my taxes. It’s the last weekend before they’re due.Oh, that’s too bad.Wait. What about your taxes?My taxes are done. I was off from work a couple of weeks ago and made hay while the sun shined. I got all of it done!I wish I would have taken advantage of my time off last week___1___All I did was lay around thehouse.And that’s all the time we have for these Words and Their Stories. But join us again next week. You can listen while you’re making dinner or riding to work. Yeah, make hay while the sun shines.1.Which of the following best matches ―make hay whilethe sun shines in paragraph 2?A.Sow nothing, reap nothing.B.Sharp tools make good work.C.Strike while the iron is hot.D.One swallow doesn’t make a summer.2.According to the underlined sentence, what feeling does the speaker express?A.AdmirableB.RegretfulC.AnnoyedD.Indifferent3.Where is the passage probably taken from?A.A radio programB.A magazineC.A brochureD.A novelBBritish sculptor Jason Taylor has made it his mission to use his talent to conserve our ecosystems by creating underwater museums. Over the years, the environmentalist has put over 850 massive artworks underwater worldwide. On February 1, 2021, Taylor launched his latest work — The Underwater Museum of Cannes.―The main goal was to bring attention to the fact that our oceans need our help,‖ Taylor told Dezeen. ―Ocean ecology has been destroyed by human activity in the Mediterranean over the past few decades, and it is not obvious what is taking place when observing the sea from afar.‖The Underwater Museum of Cannes contains 6 sculptures featuring local residents of various ages. They range from Maurice, an 80-year-old fisherman, to Anouk, a 9-year-old student. Towering over 6-feet-tall and weighing 10 tons, the faces are sectioned into two parts, with the outer part like a mask. The mask indicates that the world’s oceans appear powerful and unbeatable from the surface but house an ecosystem that is extremely fragile to careless human activities.Though the waters surrounding the sculptures now appear a pristine blue, the seabed was filled with old boat engines, pipes, and other human-made trash when the project began about four years ago. Besides removing the trash, Taylor also restored the area’s sea grass. Just one square meter of the sea grass can generate up to 10 liters of oxygen daily. The sea grass also helps prevent coastal erosion and provides habitats for many ocean creatures.―The idea of creating an underwater museum was to draw more people underwater and develop a sense of care and protection,‖ Taylor told Dezeen. ―If we threw unwanted waste near a forest, there would be a public outcry. But this is happening every day in our surrounding waters and it largely goes unnoticed.4. Why does the outer part ofthe sculptures look like a mask?A. To popularize the features of the locals.B. To remind people to protect themselves.C. To reflect people’s protection of the ocean.D. To stress the sensitiveness of the ecosystem.5. What’s paragraph 4 mainly about?A. How the project was started.B. How the sea grass was restored.C. What recovery effort the project made.D. Why the surroundings were improved.6. What can we infer from what Jason Taylor said in the last paragraph?A. The situation of the ocean is easily ignored.B. The destruction caused to the ocean is noticeable.C. Forests play a more important role in ecosystems.D. People have zero tolerance to damage done to nature.7. What might be the best title for the text?A. The Underwater Museum, a long way to go.B. The Underwater Museum, a big difference to the sea.C. The Underwater Museum, an appeal to conserve ecosystems.D. The Underwater Museum, a masterpiece of Jason Taylor.CA young female athlete in thePhilippinesrecently won many gold medals during a sports meet despite not having proper running shoes. Rhea Ballos, an 11-year-old student ofSalvationElementary Schoolin Balason,Iloilo, wasonly wearing bandages around her feet when she competed at the Iloilo Schools Sports Meet.Facebook user Valenzuela posted pictures of the girl with her feet wrapped in bandages bearing the famous Nike logo. Ballos even wrote the word “NIKE” on the sides of her “shoe” to complete the “Nike running shoes” look. The bandages were tightly wrapped around her feet, creating a thin protective layer against the track. While she was actually barefoot during the races, she was still able to defeat her competitors who all more proper footwear intended for running,According to the post, Ballos bagged the top awards in the 400-meter dash, the 800-meter run, and the 1500-meter run in the girls' categories in the inter school sporting event held in Iloilo, central Philippines.When pictures of her “Nike” footwear become popular, Flipinos on social media praised her. Many noted that instead of falling into self-pity, she was even able to make light of the situation by drawing the Nike logo on her “running shoes”. Some of the commenters of Valenzuela's post expressed how the girl deserved to be recognized by Nike and that the brand should actually give her a new pair of real Nike shoes. Others started getting in touch with the American sports brand, as well as local basketball specialty store Titan 22.It did not take long for Titan co-founder and Alaska Aces head coach Jeffrey Cariaso to take notice of Ballos' outstanding achievement. Cariaso immediately made an effort to get in touch with the young track runner. The seven-time PBA champion has since talked to the student as well as her coaches in an apparent bid to help her out.8. Why did Ballos wear bandages around her feet to compete?A. She couldn't afford to buy shoes.B. She wished to be noticed by Nike.C. She wanted to draw public's attention.D. She thought it fashionable and unique.9. What's people's attitude to Ballos' story?A. Surprised.B. Confused.C. Favorable.D. Doubtful.10. What can we infer from the last paragraph?A. Ballos will be recognized by Nike.B. Ballos will be probably helped by Cariaso.C. Ballos is bound to win more champions.D. Ballos will become a great basketball player.11. Which of the following can best describe Ballos?A. Shy and lucky.B. Kind and brave.C. Clever and outstanding.D. Gifted and optimistic.DWe've all heard it before:to be successful, get out of bed early. After all, Apple CEO Tim Cook gets up at 3:45 am, Fiat CEO Sergio Marchionne at 3:30 am and Richard Branson at 5:45 am﹣and, as we all know, "the early bird's catches the worm. "But just because some successful people wake up early, does that mean it's a trait most of them share?And if the idea of having exercised, planned your day, eaten breakfast, visualized and done one task before 8 am makesyou want to roll over and hit snooze till next Saturday, are you really doomed to a less successful life?For about half of us, this isn't really an issue. It's estimated that some 50% of the population isn't really morning or evening﹣oriented, but somewhere in the middle. Roughly one in four of us, though, tend more toward bright﹣eyed early risers, and another one in four are night owls. For them, the effects can go beyond falling asleep in front of the TV at 10 pm or being regularly late for work.Numerous studies have found that morning people are more self﹣directed and agreeable. And compared to night owls, they plan for the future more and have a better sense of well﹣being.Although morning types may achieve more academically, night owls tend to perform better on measures of memory, processing speed and cognitive(认知)ability, even when they have to perform those tasks in the morning. Night﹣time people are also more open and more creative. And one study shows that night owls areas healthy and wise as morning types﹣and a little bit wealthier.Still think the morning people sound more like CEO material?Don't set your alarm for 5 am Just yet. As it turns out, overhauling(全面改革)your sleep time may not have much effect"If people are left to their naturally preferred time, they feel much better. They say that they are much more productive. The mental capacity they have is much broader, " says Oxford University biologist Katharina Wulff. On the other hand, she says, pushing people too far out of their natural preference can be harmful. When they wake early, for example, night owls are still producing melatonin(褪黑素). "Then you disrupt it and push the body to be in the daytime mode. That can have lots of negative physiological consequence. " Wulff says, like a different sensitivity to insulin and glucose(葡萄糖)which can cause weight gain.12. What does the authordo in the first three paragraph?A. raising the problem→analyzing the problem → solving the problemB. leading in the topic→challenging a viewpoint → discussing about the topicC. presenting a viewpoint → providing supporting proofs→making a conclusionD. introducing a viewpoint →raising the question→presenting author's viewpoint13. What can we know from the 4th and 5th paragraph?A. Morning types tend to have clear goals and better mood.B. To beat night﹣time people ask them to do math calculation in themorning.C. Night owls tend to sacrifice their health for their wealth.D. Neither night owls nor morning persons perform better than the middle ones.14. Which of the following does Katharina Wulff support?A. Don't fall sleep in front of the TV.B. Avoid being regularly late for work.C. Stop setting your alarm for 5 am.D. Better not overhaul your sleep time.15. Why does the author write this article?A. To explain why some people are more successful.B.To compare the differences between early risersand night owls.C. To advise people to get up neither too early nor too late.D. To argue against this view that the Carly bird catches the worm.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。

2021-2022学年-有答案广东省深圳市某校七年级(上)月考数学试卷(11月份)

2021-2022学年-有答案广东省深圳市某校七年级(上)月考数学试卷(11月份)

2021-2022学年广东省深圳市某校七年级(上)月考数学试卷(11月份)一、选择题(每小题0分,共36分)1. −2020的倒数是()A.−2020B.2020C.12020D.−120202. 地球上的海洋面积约为361000000km2,这个数用科学记数法表示为( ) km2.A.361×106B.36.1×107C.3.61×108D.0.361×1093. 已知2a+3b=4,则整式−4a−6b+1的值是()A.5B.3C.−7D.−104. 在式子,2x+5y,0.9,−2a,−3x2y,中,单项式的个数是()A.5个B.4个C.3个D.2个5. 如图,这个几何体是由哪个图形绕虚线旋转一周形成的()A. B. C. D.6. 如图,将4×3的网格图剪去5个小正方形后,图中还剩下7个小正方形,为了使余下的部分(小正方形之间至少要有一条边相连)恰好能折成一个正方体,需要再剪去1个小正方形,则应剪去的小正方形的编号是()A.7B.6C.5D.47. 下面各组数中,相等的一组是()A.−22与(−2)2B.233与(23)3C.−|−2|与−(−2)D.(−3)3与−338. 如果规定符号“※”的意义为b※a =+1,则2※(−3)的值是( )A.8B.7C.6D.−69. 1米长的彩带,第1次剪去13,第二次剪去剩下的13,如此剪下去,剪7次后剩下的彩带长(不计损耗)为( ) A.(13)6米B.(13)7米C.(23)6米D.(23)7米10. 如图是一个正方体的表面展开图,若折叠成正方体后相对面上的两个数互为相反数,则x ,y ,z 的值分别为( )A.2,−3,−10B.−10,2,−3C.−10,−3,2D.−2,3,−1011. 如果a <0,b >0,a +b <0,那么下列关系式中正确的是( ) A.a >b >−b >−a B.a >−a >b >−b C.b >a >−b >−a D.−a >b >−b >a12. 下列说法:①倒数等于本身的数是±1;②互为相反数的两个非零数的商为−1;③如果两个数的绝对值相等,那么这两个数相等;④有理数可以分为正有理数和负有理数;⑤单项式-的系数是-,次数是6;⑥多项式3πa 3+4a 2−8是三次三项式,其中正确的个数是( ) A.2 个 B.3 个C.4 个D.5 个二、填空题(每小题3分,共12分)−8的相反数是________.已知:|x|=3,y 2=4且xy <0,则(x +y)2019=________.为鼓励节约用电,某地对居民用户用电收费标准作如下规定:每户每月用电如果不超过100度,那么每度电价按a 元收费;如果超过100度,那么超过部分每度电价按b 元收费,若某户居民在一个月内用电180度,则这个月应缴纳电费________元(用含a ,b 的代数式表示)设x 为非负实数,将x “四舍五入”到整数的值记为<x >(可读作尖括号x ),即当非负实数x 满足n −12≤x <n +12时,其中n 为整数,则<x >=n .如<0.48>=0,<5.5>=6,<3.49>=3.如果<x −2.2>=5,那么x 的取值范围是________ 三、解答题(共52分)把下列各数填在相应的大括号内15,−12,0.81,−3,14,−3.1,−4,171,0,3.14,π 正数集合{________ ...} 负数集合{________...}非负整数集合{________ ...} 有理数集合{________ ...}. 计算:(1)−9+5−(−12)+(−3);(2)−4+(−2)2×(−3);(3);(4);(5);(6).(1)由大小相同的边长为1小立方块搭成的几何体如图,请用阴影图形表示出这个几何体从不同方向看到的视图:(2)用相同形状的小立方体重新搭一几何体,使得它的俯视图和左视图与你在上图方格中所画的图一致,这样的几何体最少要________个立方块,最多要________个立方块.某自行车厂一周计划生产1400辆自行车,平均每天生产200辆,由于各种原因实际每天生产量与计划量相比有出入.下表是某周的生产情况(超产为正、减产为负):(1)根据记录可知前三天共生产________辆;(2)产量最多的一天比产量最少的一天多生产________辆;(3)该厂实行每周计件工资制,每生产一辆车可得60元,若超额完成任务,则超过部分每辆另奖15元;少生产一辆扣15元,那么该厂工人这一周的工资总额是多少?某服装厂生产一种西装和领带,西装每套定价200元,领带每条定价40元.厂方在开展促销活动期间,向客户提供两种优惠方案:①买一套西装送一条领带;②西装和领带都按定价的90%付款.现某客户要到该服装厂购买西装20套,领带x条(x>20).(1)若该客户按方案①购买,需付款________元(用含x的代数式表示);若该客户按方案②购买,需付款________元(用含x的代数式表示);(2)若x=30,通过计算说明此时按哪种方案购买较为合算?阅读下面的文字,完成后面的问题.我们知道,,…(1)那么=________;(2)用含有n(n为正整数)的式子表示你发现的规律________;(3)求式子的值.已知a是最大的负整数,b是最小的正整数,且c=a+b,请回答下列问题.(1)请直接写出a,b,c的值:a=________,b=________,c=________;(2)a,b,c在数轴上所对应的点分别为A,B,C,请在如图所示的数轴上表示出A,B,C三点;(3)在(2)的情况下.点A,B,C开始在数轴上运动,若点A和点C以每秒1个单位的速度向左运动,同时点B以每秒5个单位长度的速度向右运动,已知点B与点C之间的距离表示为BC,点A与点B之间的距离表示为AB,问:AB−BC的值是否会随着时间的变化而变化?若会,请说明理由;若不会,请求出AB−BC的值.参考答案与试题解析2021-2022学年广东省深圳市某校七年级(上)月考数学试卷(11月份)一、选择题(每小题0分,共36分)1.【答案】D【考点】倒数【解析】乘积是1的两数互为倒数.依据倒数的定义回答即可.【解答】−2020的倒数是−1,20202.【答案】C【考点】科学记数法--表示较大的数【解析】用科学记数法表示较大的数时,一般形式为a×10−n,其中1≤|a|<10,n为整数,n 的值取决于原数变成a时,小数点移动的位数,n的绝对值与小数点移动的位数相同.当原数绝对值大于1时,n是正数;当原数的绝对值小于1时,n是负数.【解答】解:361000000km2=3.61×108km2.故选C.3.【答案】C【考点】列代数式求值【解析】根据相反数的定义得:−2a−3b=−4,首先化简−4a−6b+1,然后把−2a−3b=−4代入化简后的算式,求出算式的值是多少即可.【解答】∵2a+3b=6,∴−2a−3b=−3,∴−4a−6b+7=2(−2a−8b)+1=−8+2=−7,4.【答案】C【考点】单项式【解析】直接利用单项式的定义进而分析得出答案.【解答】在式子,2x+7y,−2a2y,中,单项式有:0.8,−2a2y,共计5个.5.【答案】A【考点】点、线、面、体【解析】根据面动成体的原理:长方形绕它的一边旋转一周形成圆柱;直角三角形绕它的一直角边旋转一周形成圆锥;半圆绕它的直径旋转一周形成球;梯形绕它的一腰(垂直于底的腰)旋转一周形成圆台.【解答】根据以上分析及题目中的图形可知A旋转成圆台,B旋转成球体,C旋转成圆柱,D旋转成圆锥.6.【答案】C【考点】展开图折叠成几何体【解析】由平面图形的折叠及正方体的表面展开图的特点解题.注意只要有“田”字格的展开图都不是正方体的表面展开图.【解答】解:根据只要有“田”字格的展开图都不是正方体的表面展开图,应剪去的小正方形的编号是5.故选C.7.【答案】D【考点】有理数的乘方【解析】本题涉及负数和分数的乘方,有括号与没有括号底数不相同,对各选项计算后即可选取答案.【解答】解:A、−22=−4,(−2)2=4,故本选项错误;B、233=83,(23)3=827,故本选项错误;C、−|−2|=−2,−(−2)=2,故本选项错误;D、(−3)3=−27,−33=−27,故本选项正确.故选D.8.【答案】A【考点】有理数的混合运算【解析】原式利用题中的新定义计算即可求出值.【解答】根据题中的新定义得:原式=+7=+1=+1=7+6=8.9.【答案】D【考点】有理数的乘方【解析】根据题意表示出各次剩下的米数,依此类推得到第7次剩下的即可.【解答】解:第1次剩下1−13=23米;第2次剩下23×(1−13)=(23)2米;…,依此类推,剪7次剩下的彩带长为(23)7米.故选D10.【答案】B【考点】相反数正方体相对两个面上的文字【解析】正方体的对面不存在公共部分可确定出对面,然后可得到x、y、z的值.【解答】x与10为对面,y与−2为对面,z与3为对面,∴x=−10,y=2,z=−3,11.【答案】D【考点】不等式的性质【解析】先确定a,b的符号与绝对值,进而放到数轴上判断4个数的大小即可.【解答】∵a<0,b>0∴−a>0−b<0∵a+b<0∴负数a的绝对值较大∴−a>b>−b>a.12.【答案】B【考点】倒数有理数的概念及分类多项式单项式绝对值相反数有理数的除法【解析】根据倒数的定义、相反数定义、绝对值的性质、有理数的分类、单项式的系数和次数定义、多项式的次数和项的定义逐个判断即可.【解答】倒数等于本身的数是±1,故①正确;互为相反数的两个非零数的商为−1,故②正确;如果两个数的绝对值相等,那么这两个数相等或互为相反数;有理数可以分为正有理数和负有理数、6,故④错误;单项式-的系数是-π,故⑤错误;多项式3πa3+4a2−8是三次三项式,故⑥正确;即正确的个数有5个,二、填空题(每小题3分,共12分)【答案】8【考点】相反数 【解析】一个数的相反数就是在这个数前面添上“-”号. 【解答】 −8的相反数是8. 【答案】 ±1【考点】有理数的混合运算 【解析】先根据绝对值的性质和有理数的乘方,求出x 、y 的值,然后根据x ⋅y <0,进一步确定x 、y 的值,再代值求解即可. 【解答】∵ |x|=3,y 2=4, ∴ x =±3,y =±2, ∵ x ⋅y <7, ∴ x 与y 异号,∴ x =3时,y =−2,(x +y)2019=4; x =−3时,y =2,(x +y)2019=−4. ∴ (x +y)2019=±1; 【答案】 (100a +80b) 【考点】 列代数式 【解析】因为180>100,所以其中100度是每度电价按a 元收费,多出来的80度是每度电价按b 元收费. 【解答】100a +(180−100)b =100a +80b . 【答案】 6.7≤x <7.7 【考点】近似数和有效数字 【解析】利用对非负实数x “四舍五入”到整数的值记为<x >,进而得出x 的取值范围.. 【解答】∵ <x −2.2>=5, ∴ 4.5≤x −2.2<5.5 ∴ 6.7≤x <7.7. 三、解答题(共52分) 【答案】15,0.81,14,171,3.14,π,−12,−3,−3.1,−4,15,171,0,15,−12,0.81,−3,14,−3.1,−4,171,0,3.14 【考点】有理数的概念【解析】根据正数,负数,非负整数,有理数的定义可得出答案.【解答】解:正数集合{15, 0.81, 14, 171, 3.14, π ...}负数集合{−12, −3, −3.1, −4...}非负整数集合{15, 171, 0 ...}有理数集合{15, −12, 0.81, −3, 14, −3.1, −4, 171, 0, 3.14...}.故答案为:{15, 0.81, 14, 171, 3.14, π ...};{−12, −3, −3.1, −4...};{15, 171, 0 ...};{15, −12, 0.81, −3, 14, −3.1, −4, 171, 0, 3.14...}.【答案】−9+5−(−12)+(−3)=−9+5+12−8=−12+17=5;−4+(−3)2×(−3)=−5+4×(−3)=−6−12=−16;=7−9=−2;=×(−36)+=−28−14+27=−15;=-×(−9×=-×(−8−4)=-×(−8)=6;=−6×−7×=(−5−7+12)×=0×=0.【考点】有理数的混合运算【解析】(1)先化简,再计算加减法;(2)先算乘方,再算乘发,最后算加法;(3)先算乘除,后算加法;(4)根据乘法分配律计算;(5)先算乘方,再算乘法,最后算加法;如果有括号,要先做括号内的运算;(6)将除法变为乘法,再根据乘法分配律计算.【解答】−9+5−(−12)+(−3)=−9+5+12−8=−12+17=5;−4+(−3)2×(−3)=−5+4×(−3)=−6−12=−16;=7−9=−2;=×(−36)+=−28−14+27=−15;=-×(−9×=-×(−8−4)=-×(−8)=6;=−6×−7×=(−5−7+12)×=0×=0.【答案】;5,7【考点】简单组合体的三视图作图-三视图【解析】(1)左视图有3列,每列小正方形数目分别为1,2,1;俯视图有2列,每列小正方形数目分别为2,1,依此画出图形即可;(2)由俯视图易得最底层小立方块的个数,由左视图找到其余层数里最少个数和最多个数相加即可.【解答】如图所示:;由俯视图易得最底层有4个小立方块,第二层最少有1个小立方块;第二层最多有6个小立方块,所以最多有7个小立方块.故答案为:5,5.【答案】5992684675【考点】正数和负数的识别【解析】(1)根据有理数的加法,可得答案;(2)根据最大数减最小数,可得答案;(3)根据实际生产的量乘以单价,可得工资,根据超出的部分或不足的部分乘以每辆的奖金,可得奖金,根据工资加奖金,可得答案.【解答】解:(1)5−2−4+200×3=599(辆);(2)16−(−10)=26(辆);(3)5−2−4+13−10+16−9=9,(1400+9)×60+9×15=84675(元).【答案】40x+3200,3600+36x(2)当x=30元时,方案①需付款为:40x+3200=40×30+3200=4400(元),方案②需付款为:3600+36x=3600+36×30=4680(元),∵4400<4680,∴选择方案①购买较为合算.【考点】列代数式求值列代数式【解析】(1)方案①需付费为:西装总价钱+20条以外的领带的价钱,方案②需付费为:西装和领带的总价钱×90%;(2)把x=30代入(1)中的两个式子算出结果,比较即可.【解答】解:(1)方案①需付款为:200×20+(x−20)×40=40x+3200(元);方案②需付款为:(200×20+40x)×0.9=3600+36x(元).故答案为:40x+3200;3600+36x.(2)当x=30元时,方案①需付款为:40x+3200=40×30+3200=4400(元),方案②需付款为:3600+36x=3600+36×30=4680(元),∵4400<4680,∴选择方案①购买较为合算.【答案】=1−+…+=8−=.【考点】有理数的混合运算列代数式【解析】(1)根据题目中的式子,可以将所求式子分解;(2)根据题目中式子的特点,可以写出第n个等式;(3)根据题目中的式子,先裂项,然后计算即可解答本题.【解答】=,故答案为:;第n个式子为,故答案为:;=1−+…+=8−=.【答案】−1,1,0(2)如图:(3)不会.假设运动时间为t,则BC=(1+5t)−(0−t)=1+6t,AB=(1+5t)−(−1−t)=2+6t,∴AB−BC=2+6t−(1+6t)=1,∴AB−BC的值不会随着时间的变化而变化,AB−BC的值为1.【考点】数轴实数列代数式求值【解析】(1)根据题意可得;(2)在数轴上直接标出;(3)先求出AB,BC的值,再计算AB−BC的值,可得AB−BC的值是定值.【解答】解:(1)由题意可得a=−1,b=1,c=−1+1=0.故答案为:−1;1;0.(2)如图:(3)不会.假设运动时间为t,则BC=(1+5t)−(0−t)=1+6t,AB=(1+5t)−(−1−t)=2+6t,∴AB−BC=2+6t−(1+6t)=1,∴AB−BC的值不会随着时间的变化而变化,AB−BC的值为1.。

2021年深圳宝安第一外国语学校七年级入学分班考试数学试卷及答案解析

2021年深圳宝安第一外国语学校七年级入学分班考试数学试卷及答案解析

2021年深圳宝安第一外国语学校七年级入学分班考试数学试卷一、你一定能选对!(本大题共有10小题,每小题3分,共30分)下列各题均有四个备选答案,其中有且只有一个是正确的,请将正确答案的代号在答题卡上将对应的答案标号涂黑。

1.中国人很早开始使用负数,中国古代数学著作《九章算术》的“方程”一章,在世界数学史上首次正式引入负数.如果收入100元记作+100元,那么﹣60元表示()A.收入60元B.收入40元C.支出60元D.支出40元2.﹣3的倒数是()A.3B.﹣3C.13D.−133.2021年5月22日,我国自主研发的“祝融号”火星车成功到达火星表面.已知火星与地球的最近距离约为55000000千米,数据55000000用科学记数法表示为()A.55×106B.5.5×107C.5.5×108D.0.55×108 4.单项式−25a3b的系数与次数分别是()A.−25,3B.25,4C.−25,4D.﹣2,35.下列运算中,正确的是()A.3a+2b=5ab B.5a3﹣4a2=aC.﹣x2+3x2=2x2D.4x2y﹣3y2x=x2y6.下列各组数中,互为相反数的是()A.24与(﹣2)4B.﹣(﹣2)与﹣|﹣2|C.(﹣2)3与﹣23D.(﹣1)4与﹣1×47.有理数a,b在数轴上对应的点的位置如图所示,则下列结论错误的是()A.|a|<|b|B.﹣a<bC.a﹣b<0D.(a+1)(b﹣1)<08.一种商品每件成本为a元,原来按成本增加40%定出售价,现在由于库存积压减价,打八折出售,则每件盈利()元.A.0.1a B.0.12a C.0.15a D.0.2a9.有理数a,b在数轴上对应的点分别为A、B,要使算式﹣12﹣|a□b|计算出来的值最大,则在“□”所在位置,填入的运算符号为()A.+B.﹣C.×D.÷10.如图,将形状大小完全相同的★按照一定规律摆成下列图形,第1幅图中★的个数为a1,第2幅图中★的个数为a2,第3幅图中★的个数为a3…,以此类推,第n幅图中★的个数为a n,则1+2+3+⋯+2021的值为()A.20202021B.20202021C.20212022D.20212022二、填空题(本大题共有6小题,每小题3分,共18分)下列各题不需要写出解答过程,请将结论直接填写在答题卷的指定位置11.2021的相反数为.12.用四舍五入法取近似数:1.2068≈(精确到0.01).13.多项式ax2﹣y+3xy4﹣5是次项式,常数项是.14.已知m为最大的负整数,x−12y=−12,则(y﹣2x)2021+m2020=.15.下列说法:①若ab<0,则|b﹣a|=|b|+|a|;②若a3=b3则a2=b2;③两个四次多项式的和一定是四次多项式;④多项式x2﹣3kxy﹣3y2+13xy﹣8合并同类项后不含xy项,则k的值是19.其中一定正确的是.(填序号)16.如图,王明家的住房平面图是一个长方形,被分割成3个正方形和2个长方形,其中标号相同的两个图形形状大小一样,若原住房平面图(长方形)的周长为m,则标号为②的正方形边长为.三、解下列各题(本大题共8小题,共72分)下列各题需要在答题卷的指定位置写出文字说明、证明过程、演算步骤或画出图形。

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2020-2021学年广东省深圳市宝安区新安中学(集团)外国语学校七上月考数学
试卷
一、选择题(共12小题;共60分)
1. 在 −(−8),−∣−7∣,−0,(−23)2
这四个数中,负数有 ( )
A. 1 个
B. 2 个
C. 3 个
D. 4 个
2. 1
4 的倒数是 ( )
A. −4
B. 4
C. 1
4
D. −1
4
3. 下列有理数大小关系判断正确的是 ( )
A. ∣−3∣<∣+3∣
B. 0>∣−10∣
C. −(−19)>−∣∣−110∣∣
D. −1>−0.01
4. (−2)3 的值是 ( )
A. −6
B. 6
C. 8
D. −8
5. 把 (+5)−(+3)−(−1)+(−5) 写成省略括号后的算式为 ( )
A. −5−3+1−5
B. 5−3−1−5
C. 5+3+1−5
D. 5−3+1−5 6. 如果有理数 a ,b 满足条件 ab >0,那么 a ÷b 的值是 ( )
A. 正数
B. 负数
C. 非负数
D. 非正数 7. 下列说法正确的是 ( )
A. 所有的整数都是正数
B. 不是正数的数一定是负数
C. 0 是最小的有理数
D. 整数和分数统称有理数
8. 计算 (−36)×(1
3−3
4+1
6
) 时,可以使运算简便的是 ( )
A. 乘法交换律
B. 乘法分配律
C. 加法结合律
D. 乘法结合律
9. 数 1,0,−2
3,−2 中最大的是 ( )
A. 1
B. 0
C. −2
3 D. −2
10. 计算 ∣−1∣−3,结果正确的是 ( )
A. −4
B. −3
C. −2
D. −1
11. 近年来,国家重视精准扶贫,收效显著,据统计,约有 65000000 人脱贫.将 65000000 用科学
记数法表示应为 ( )
A. 6.5×107
B. 65×106
C. 0.65×107
D. 6.5×108
12. 若规定“!”是一种数学运算符号,且 1!=1,2!=2×1=2,3!=3×2×1=6,⋯,则 100!
98! 的
值为 ( )
A. 50
49
B. 99
C. 9900
D. 2
二、填空题(共4小题;共20分) 13. 我市某天的最高气温是 4∘C ,最低气温是 −1∘C ,则这天的日温差是 ∘
C .
14. 11.绝对值与倒数均等于它本身的数是 . 15. 已知 ∣a −2∣+(b +3)2=0,则 b a = .
16. (−2)×(−2)×(−2)×⋯×(−2)×(−2)⏟
共有2005个(−2)
的计算结果,用以 2 为底的幂的形式表示是 .
三、解答题(共7小题;共91分) 17. 计算:
(1)12−(−18). (2)−32+∣−18∣.
18. 有 20 筐白菜,以每筐 25 kg 为标准,超过或不足的千克数分别用正、负数来表示,记录如下:
与标准质量的差值/kg −3−2−1.501 2.5
筐数
1
8
2
32
4
(1)20 筐白菜中,最重的一筐比最轻的一筐重多少千克? (2)与标准质量相比,20 筐白菜总计超过或不足多少千克? (3)若白菜每千克售价 2.6 元,则出售这 20 筐白菜可卖多少元?
19. 把下列各数分别填在相应的大括号内:25,−0.91,π,3.14,−7,0,−50,7
8,9.
(1)整数有:{ }; (2)分数有:{ }; (3)正整数有:{ }; (4)负整数有:{ }; (5)正分数有:{ }; (6)负分数有:{ };
20. 已知:数轴上 A ,B 两点表示的有理数分别为 a ,b ,且 (a −1)2+∣b +2∣=0.
(1)求 (a +b )2015 的值. (2)数轴上的点 C 分别与 A ,B 两点的距离的和为 7,求点 C 在数轴上表示的数 c 的值.
21. 计算:
(1)16+(−25)+24+(−35); (2)(−3
4)×(−11
2)÷(−21
4);
(3)23×(−5)−(−3)÷3
128; (4)∣−10∣+∣(−4)2−(1−32)×2∣.
22. 已知 ∣x ∣=3,∣y ∣=7.
(1)写出 x 和 y 的值; (2)若 x <y ,求 x −y 的值. 23. 阅读 (−1
12)÷(2
3−1
4+5
6
) 的计算方法.
解:设原式的商为 x , 因为 x =0,
所以 1
x =(2
3−1
4+5
6)÷(−1
12)=(2
3−1
4+5
6)×(−12)=−8+3−10=−15, 所以商 x =−1
15.
请按以上方法计算 (1
42)÷(7
6−2
3−3
14).
答案第一部分
1. A 【解析】−(−8)=8;−∣−7∣=−7;−0=0;(−2
3)
2
=4
9

∴负数有1个.
2. B
3. C
4. D
5. D
6. A
7. D 【解析】A、负整数和0就不是正数,显然A错误;
B、不是正数,有可能是零,所以B错误;
C、负有理数比零小,错误;
D、正确.
8. B
9. A
10. C
11. A
12. C
第二部分
13. 5
14. 1
【解析】绝对值与倒数均等于它本身的数是1.
故答案为:1.
15. 9
16. −22005
第三部分
17. (1)原式=2+18=30.
(2)原式=−9+18=9.
18. (1)最重的一筐比最轻的一筐重2.5−(−3)=5.5(kg).
(2)−3×1+(−2)×8+(−1.5)×2+0×3+1×2+2.5×4=−10(kg).答:与标准质量相比,20筐白菜总计不足10kg.
(3)2.6×(25×20−10)=1274(元).
答:出售这20筐白菜可卖1274元.
19. (1)25,−7,0,−50,9
(2)−0.91,3.14,7
8
(3)25,9
(4)−7,−50
(5) 3.14,7
8
(6) −0.91
20. (1) ∵ (a −1)2+∣b +2∣=0, ∴ a −1=0,b +2=0, 解得 a =1,b =−2,
∴ (a +b )2015=(1−2)2015=(−1)2015=−1;
(2) ∵ a =1,b =−2,数轴上 A ,B 两点表示的有理数分别为 a ,b ,数轴上的点 C 与 A ,B 两点的距离的和为 7,
∴ 点 C 可能在点 B 的左侧或点 C 可能在点 A 的右侧, 当点 C 在点 B 的左侧时,1−c +(−2−c )=7,得 c =−4, 当点 C 在点 A 的右侧时,c −1+c −(−2)=7,得 c =3, 即点 C 在数轴上表示的数 c 的值是 −4 或 3.
21. (1) 原式=−(16+24)+[(−25)+(−35)]=40+(−60)=−20. (2) 原式=(−3
4)×(−3
2)÷(−9
4)=9
8×(−4
9)=−1
2

(3)
原式=23×(−5)−(−3)×
1283
=(−115)−(−128)=(−115)+128=13.
(4) 原式=10+∣16−(−8)×2∣
=10+32
=42.
22. (1) ∵∣x ∣=3,∣y ∣=7, ∴x =±3,y =±7.
(2) 当 x <y 时,x =3,y =7 或 x =−3,y =7, 此时 x −y =−4或−10. 23. 设原式的商为 y , ∵y ≠0, ∴
1y
=(76
−2
3−
314
)÷142=(76−23−3
14)×42=49−28−9=12, ∴ 商 y =112
.。

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