2022-2023学年江苏省南京市联合体七年级第二学期期末数学练习试卷及评分标准

南京市联合体七年级第二学期期末数学试卷
一、选择题（本大题共6小题，每小题2分，共12分） 1．计算(－a 2)3的结果是（ ）
A ．a 5
B ．－a 5
C ．a 6
D ．－a 6
2．下列运算正确的是（ ）
A ．a 3＋a 2＝a 5
B ．a 3·a 2＝a 6
C ．a 3÷a 2＝a
D ．(a 3)2＝a 9
3．若x ＞y ，则下列各式不正确的是（ ）
A ．x ＋2＞y ＋2
B ．x －2＞y －2
C ．x 2＞y 2
D ．－2x ＞－2y
4．如图，l 1∥l 2，若∠1＝115°，则∠2的度数为（ ）
5．已知M ＝x 2＋x ，N ＝3x －1，则M ，N 的大小关系是（ ）
A ．M ≥N
B ．M ＞N
C ．M ≤N
D ．M ＜N
6．如图，两面镜子AB ，BC 的夹角为∠α，当光线经过镜子后反射，∠1＝∠2，∠3＝∠4．若∠α＝70°，则∠β的度数是（ ）
A ．30°
B ．35°
C ．40°
D ．45°
二、填空题（本大题共10小题，每小题2分，共20分）
7．某种花粉颗粒的直径约为0.000 031m ，将0.000 031用科学记数法表示为 ． 8．若x 2－6x ＋m （m 为常数）是一个完全平方式，则m 的值是 ． 9．若一个多边形的每个内角都为144°，则这个多边形的边数是 ． 10．若(a ＋b )2＝7，a 2＋b 2＝3，则ab ＝ ．
11．已知方程组⎩
⎪⎨⎪⎧3x ＋y ＝5，x ＋3y ＝－1．则x ＋y ＝ ．
12．一个直角三角形的两个锐角的差是20°，则最小的锐角的度数为 °． 13．已知y ＝2x ＋1，若－1＜y ≤3，则x 的取值范围是 ．
A ．55°
B ．65°
C ．75°
D ．85°
l 1
l 2
1
2
（第4题）
（第6题） A
14．如图，△ABC 是直角三角形．若l 1∥l 2，则∠1－∠2＝ °．
15．若关于x 的一元一次不等式x ＋1≤m 只有1个正整数解，则m 的取值范围是 ． 16．如图，四边形ABOC 中，∠BAC 与∠BOC 的角平分线相交于点P ，若∠B ＝16°，∠C ＝42°，则∠
P ＝ °．
三、解答题（本大题共10小题，共68分．解答时应写出文字说明、证明过程或演算步骤） 17．（6分）计算：
（1）20＋23－2－3
； （2）(x －3y )(2x ＋y )．
18．（6分）因式分解：
（1）2a 3－12a 2＋18a ； （2）(a －3)2－16．
A
C
B
l 1
1
2 l 2 （第14题）
A
B
O
P （第16题）
19．（6分）先化简，再求值：(x －2y )2－(x ＋2y )(x －2y )，其中x ＝－1，y ＝1
2．
20．（6分）解方程组⎩⎨⎧3x －y ＝7，
x ＋3y ＝－1．
21．（6分）解不等式组⎩⎨⎧2(x －1)≤－4，
x 3－x －12
＜1．并写出它的整数解．
22．（6分）比较大小．
（1）当a ＞1时，a
a ＋1
2
（填“＞”、“＜”或“＝”）； （2）说明第（1）小题的正确性．
23．（6分）如图，平移线段AB ，使点A 移动到点A '的位置．
（1）尺规作图，保留作图痕迹； （2）作图的依据是 ．
24．（8分）如图， D 、E 、F 、G 是△ABC 边上的点，DE ∥BC ，∠1＝∠2．
求证：DG ∥FC ．
A
E G C
B
D
F
1
2
（第24题）
（第23题）
A
A '
25．（8分）某校计划购买A 型和B 型两种笔记本作为奖品发放给学生，若购买A 型笔记本5本，B
型笔记本8本，共需80元；若购买A 型笔记本15本，B 型笔记本4本，共需140元． （1）A 型和B 型笔记本每本的价格分别是多少元？
（2）该校计划购买A 型和B 型两种笔记本共80本，费用不超过500元，A 型笔记本最多买多少
本？
26．（10分） 【初步认识】
（1）如图①，在△ABC 中，BP ，CP 分别平分∠ABC ，∠ACB ．
求证：∠BPC ＝90°＋1
2
∠A ．
A
B
C
P
①
【继续探索】
（2）如图②，在△ABC 中，BM 平分∠ABC ，CM 平分△ABC 外角∠ACD ．
求证：∠M ＝1
2
∠A ．
（3）如图③，BN 、CN 分别平分△ABC 外角∠EBC ，∠FCB ．
则∠N 与∠A 的数量关系是 ．
（4）如图④，△ABC 中的两内角平分线交于P 点，两外角平分线交于N 点，一内角平分线与一外
角平分线交于M 点．设∠BPC ＝a °，∠M ＝b °，∠N ＝c °，则a ，b ，c 之间的关系是 ．
A
B
C
M
② D
A
B
C
N ③
E
F
A
B
C
N ④
E
F
M
P
D
七年级数学参考答案
一、选择题（本大题共6小题，每小题2分，共12分）
二、填空题（本大题共10小题，每小题2分，共20分）
7．3.1×10－
5； 8．9； 9．10； 10．2； 11．1；
12．35； 13．－1＜x ≤1； 14．90； 15． 2 ≤m ＜3； 16．13．
三、解答题（本大题共68分．请在答题卡指定区域．．．．．．．
内作答，解答时应写出文字说明、证明过程或演算步骤）
17．（6分）（1）原式＝1＋8－1
8
， ·········································································· 2分
＝
718
． ··················································································· 3分 （2）原式＝2x 2＋xy －6xy －3y 2， ······················································· 2分
＝2x 2－5xy －3y 2． ······························································· 3分
18．（6分）（1）原式＝2a (a 2－6a ＋9)， ································································· 1分
＝2a (a －3)2， ········································································· 3分
（2）原式＝[(a －3)＋4][( a －3)－4]， ················································· 1分
＝(a ＋1)(a －7)． ·································································· 3分
19．（6分）原式＝x 2－4xy ＋4y 2－(x 2－4y 2)， ························································ 2分
＝8y 2－4xy ， ·················································································· 4分 把x ＝－1，y ＝1
2
代入得，原式＝4． ·················································· 6分
20．（6分）解法一：
由①得y ＝3x －7……③， ····································································· 1分 把③代入②得x ＝2， ············································································ 3分
所以原方程组的解为⎩⎨⎧x ＝2，y ＝－1．
···························································· 6分
解法二：
①×3得9x －3y ＝21……③， ································································ 1分 ②＋③得x ＝2， ···················································································· 3分 把x ＝2代入①得y ＝－1， ··································································· 5分
所以原方程组的解为⎩⎨⎧x ＝2，y ＝－1．
···························································· 6分
21．（6分）由①得x ≤－1， ··················································································· 2分
由②得x ＞－3， ···················································································· 4分 ∴不等式组的解集是－3＜x ≤－1，····················································· 5分 ∴不等式组的整数解是：－1，－2． ·················································· 6分
22．（6分）（1）＞． ······························································································· 2分
（2）∵a ＞1，
∴a ＋a ＞a ＋1， ··········································································· 4分 ∴2a ＞a ＋1， ··············································································· 5分 ∴a ＞a ＋1
2
． ················································································· 6分
23．（6分）（1）
······································································································ 4分 （2）一个图形和它经过平移所得的图形中，两组对应点连线平行且相等． ··················································································································· 6分
24．（8分）∵DE ∥BC ，
∵∠ADG ＝∠ADE ＋∠1，∠AFC ＝∠B ＋∠2, ∠1＝∠2， ············ 4分 ∴∠ADG ＝∠AFC ， ··········································································· 6分 ∴DG ∥FC ． ························································································ 8分
25．（8分）（1）设A 型笔记本每本x 元，B 型笔记本每本y 元，
根据题意得⎩⎨⎧5x ＋8y ＝80，
15x ＋4y ＝140．
····················································· 3分
解得⎩⎨
⎧x ＝8，y ＝5．
答：A 型笔记本每本8元，B 型笔记本每本5元． ················· 5分 （2）设购买A 型笔记本m 本，
根据题意得8m ＋5(80－m )≤500， ··········································· 7分 解得m ≤1003 ，
∴m 最大取33，
答：A 型笔记本最多买33本． ················································· 8分
26．（10分）（1）∵BP ，CP 分别平分∠ABC ，∠ACB ，
∴∠PBC ＝12∠ABC ，∠PCB ＝1
2∠ACB ． ································· 1分
∴∠BPC ＝180°－∠PBC －∠PCB ＝180°－1
2
(∠ABC ＋∠ACB )
＝180°－12(180°－∠A )＝90°＋1
2
∠A ． ················· 3分
（2）∵∠ACD 是△ABC 的外角，∠MCD 是△MBC 的外角，
∴∠ACD ＝∠ABC ＋∠A ，∠MCD ＝∠MBC ＋∠M ． ·············· 4分 ∴∠M ＝∠MCD －∠MBC ＝12(∠ACD －∠ABC )＝1
2∠A ． ······· 6分
（3）∠N ＝90°－1
2∠A ． ·································································· 7分
（4）a －c ＝2b ． ··············································································· 10分。