新人教版2018年七年级暑假数学自我强化巩固训练试卷

2018年七年级暑假数学自我强化巩固训练试卷

一、填空题

1．计算(﹣2＋1)＝ ▲ ，3＝ ▲ ．

2．一种花瓣的花粉颗粒直径约为0.0000079米，0.0000079用科学记数法表示为 ▲ ． 3．若xm=3，xn=2，则xmn＝ ▲ ．

－

0－2

4．命题“等角的补角相等”的逆命题是 ▲ ．

A ▲ °5．如图∠1，∠2，∠3是五边形ABCDE的外角，且∠1＝∠2＝∠3＝72°，则∠C＋∠D＝．

D C 3 B 2 A A

D ′

E D

EF65B

C （第6题）

′

F

C

BDCE 1 （第5题）

（第9题）

6．如图，一个长方形纸片沿EF折叠后，点D、C分别落在D′、C′的位置，若∠EFB＝65o，则∠AED′等于 ▲ °．

7．已知：a＋b＝3，ab＝﹣2，则a2b＋ab2的结果是 ▲ ．

?x＝－1，?3x＋2y＝m，

8．已知?是二元一次方程组?的解，则m－n的值是 ▲ ．

?y＝2．?nx－y＝1．

9．如图，D为△ABC的BC边上任意一点，F为AD的中点，E为BF的中点，△CEF的面积为5，则△ABC的面积为 ▲ ．

10．如果关于x的不等式2x－4m≤0有3个正整数解，则 m 的取值范围是 ▲ ． 一、选择题

11．下列计算正确的是

A．a3＋a2＝a5

B. a3－a2＝a

C. a3·a2＝a 6

D.a3÷a2＝a

12．若一个三角形的两边长分别为3cm，7cm，则第三边长可能是

A．3cm

B．4cm

C．7cm

D．10cm

13．不等式-2x≥-4的解集在数轴上表示为

A． B． C． D． 14．下列命题中，属于真命题的是

A．相等的角是对顶角 C．若a＝b，则a＝b

B．同旁内角互补

D．如果直线l1∥l2，直线l2∥l3，那么l1∥l3．

15．如图，将一副直角三角板如图所示放置，使含30°角的三角板的一条直角边和含45°角的三角板的一条直角边重合，则∠1的度数为 A．75

（第15题）

E 1

B．60°

C．45°

A 1 3 4 C D

D．80°

222333B 2 （第16题）

（第18题）

16．如图，给出下列条件：①∠1=∠4；②∠2=∠3；③AB∥CD，且∠A＝∠C；④∠EBC＝∠A；其中能推出AD∥BC的条件有

A．①②③

B．①③④

C．①②④

D．②③④

17．小君问叔叔的年龄，叔叔说：“我像你这么大时，你才4岁，你到我这么大时，我就40岁了．”小君和叔叔现在的年龄分别是 A．8岁、20岁

B．16岁、28岁

C．15岁、27岁

D．9岁、21岁

18．有5张边长为2的正方形纸片，4张边长分别为2、3的矩形纸片，6张边长为3的正方形纸片，从其中取出若干张纸片，且每种纸片至少取一张，把取出的这些纸片拼成一个正方形（按原纸张进行无空隙、无重叠拼接），则拼成正方形的边长最大为 A．6 三、解答题 19．（6分）计算：

（1）(－2a3)2－a2·(－a4)－a8÷a2； （2）（a+2b）（a+b）－3a（a+b）．

20．（6分）因式分解：

（1）ab2－6ab＋9a； （2）a2(x－y)－x＋y．

B．7

C．8

D．10

??2x－y＝2，

21．（5分）解方程组?

?4x－3y＝8．?

x－4＜3(x－2)，??

22．（5分）解不等式组 ?2x＋1，并求出最小整数解．

＋1＜x．??3

23. （7分）叙述并证明三角形内角和定理．

定理： ． 已知： ． 求证： ． 证明：

24. （8分）如图，已知AB∥CD，∠1＝∠F，∠2＝∠E.

（1）若∠E＝40°，求∠F；

（2）思考：∠E与∠F有怎样的数量关系，并说明理由．

25．（8分）某工程队正进行某项工程的施工，有大量的沙石需要运输，该工程队车队现有载重量为10吨、15吨的两种卡车共10辆，一次能运输130吨沙石． （1）求该车队现有载重量为10吨、15吨的卡车各有多少辆？

（2）随着工程的进展，该车队需要一次运输沙石的能力不低于270吨，准备新增购这两．种卡车共10辆（两种卡车都要有），请帮车队设计合理的购买方案． ．．．

26. （9分）如图，已知AM∥BN，∠A＝m，点P是射线AM上一动点(与点A不重合)，BC、BD分别平分∠ABP和∠PBN，分别交射线AM于点C、D.，且满足|∠CBP－∠PBD|＝n． （1）若m＝60°，求∠CBD的度数； （2）若m＝60°，n＝0°，求∠ABD的度数；

（3）求∠CBP、∠PBD的度数．（用含m、n的代数式表示）

27.（10分）【初步感知】

定义：两条相交直线的夹角的角平分线所在的直线叫做相交线的和谐线．

如图①，直线AB与直线CD相交于点O，l1是∠AOD的角平分线所在的直线， l2是∠AOC的角平分线所在的直线，则l1与l2就是直线AB、CD的和谐线． （1）直线AB、CD的两条和谐线的位置关系为 ；

【问题解决】

如图②，已知a∥b，直线c与直线a、b交于点A、B，

（2）直线a、c的和谐线与直线b、c的和谐线有怎样的位置关系，并说明理由． c c

图②

图③

B

b

B β

A a

A

α

a

图①

C

B

A l1

D O l2

b

【延伸推广】

如图③，已知直线c与直线a、b交于点A、B，若a、c的夹角为α，b、c的夹角为β， α＋β≠180°，a、c的和谐线与b、c的和谐线交于点C．

（3）画出图形，直接写出∠ACB的度数．（用含α、β的代数式表示）

参考答案

一、填空题（每小题2分，共计20分）

13－

1．1， 2．7.9×106 3． 4．如果两个角的补角相等，那么这两个角相等

923

5．216 6．50 7．－6 8．4 9．20 10．≤m＜2

2二、选择题（每小题2分，共计16分）

题号 答案 11 D 12 C 13 C 14 D 15 A 16 D 17 B 18 C 三、解答题（本大题共9小题，共计64分） 19．(6分)计算：

（1）解：原式＝4a6＋a6－a6 ······································································· 2分 ＝4a6． ······················································································ 3分 （2）解：原式＝a2+ab+2ab+2b2﹣3a2﹣3ab ···················································· 2分

＝－2a2+2b2 ················································································· 3分

20．(6分)

（1）解：原式＝a（b2－6b＋9) ··································································· 2分

＝a(b－3)2 ···················································································· 3分

（2）解：原式＝a2(x－y)－(x－y) ································································· 1分

＝(x－y)(a2－1) ················································································ 2分 ＝(x－y) (a＋1) (a－1) ······································································· 3分

21．(5分)

解：由②－①×2，得－y=4，

解得：y=－4， ························································································· 2分 把y=－4代入①，得x=－1， ······································································ 4分

?x＝－1，∴原方程组的解为? ······································································· 5分

?y＝－4．

22．(5分)

解：由①，得x＞1， ················································································· 1分

由②，得x＞4， ················································································· 3分 ∴不等式组的解集是x＞4， ········································································ 4分

∴最小整数解是5． ·················································································· 5分 23．(7分)

解：定理：三角形的内角和是180°；

已知：如图，△ABC的三个内角分别为∠A、∠B、∠C；

求证：∠A+∠B+∠C=180°． ······································································· 3分

············································································································· 4分 证明：过点A作直线MN，使MN∥BC．

∵MN∥BC，∴∠B=∠MAB，∠C=∠NAC．（两直线平行，内错角相等） ············· 5分 ∵∠MAB+∠NAC+∠BAC=180°，（平角定义）

∴∠B+∠C+∠BAC=180°，（等量代换） ························································· 6分 即∠A+∠B+∠C=180°． ············································································· 7分 24．(8分)

（1）解： ∵∠E＝40°，∠2＝∠E，∴∠2＝40°．…………1分

∴∠DCE＝180°－（∠E+∠2）＝100°．…………2分 ∵AB∥CD，∴∠ABF+∠DCE＝180°．

∴∠ABF＝180°－100°＝80°． ……………………3分 ∵∠1＝∠F，

11

∴∠F＝（180°－∠ABF）＝（180°－80°）＝50°．

22

······························································································ 4分

（2）解： ∠E+∠F＝90°．

EBGCA12DOHF（第24题）

∵∠2＝∠E，∠1＝∠F，

11

∴∠E＝（180°－∠DCE），∠F＝（180°－∠ABF）． ····················· 6分

2211

∴∠E+∠F＝（180°－∠DCE）+（180°－∠ABF）

22

1

＝（360°－∠DCE－∠ABF）． ····································· 7分 2

∵AB∥CD，∴∠ABF+∠DCE＝180°，

1

∴∠E+∠F＝（360°－180°）＝90°． ············································ 8分

2

25．(8分)

解：（1）设该车队现有载重量为10吨、15吨的卡车各有x辆和y辆．

?x＋y＝10?x＝4

根据题意，得：?，解得：?． ······························ 3分

?10x＋15y＝130?y＝6

答：该车队现有载重量为10吨、15吨的卡车各有4辆和6辆． ··········· 4分 （2）设新增购载重量为10吨的卡车z辆，则新增购15吨的卡车（10－z）辆．

根据题意，得10（4+z）+15（6+10－z）≥270，解得z≤2．···················· 6分 则z的正整数解为1，2．

答：新增载重量为10吨的卡车1辆或2辆，新增15吨的卡车9辆或8辆． ··················································································································· 8分 26．(9分)

解（1）∵AM∥BN，∴∠A+∠ABN=180°．

∵∠A＝m＝60°，∴∠ABN=120°． ······················································· 1分 ∵BC、BD分别平分∠ABP和∠PBN， 11

∴∠CBP＝∠ABP、∠PBD＝∠PBN，

22

111

∴∠CBD＝∠ABP+∠PBN＝∠ABN＝60°． ········································· 3分

222（2）∵|∠CBP－∠PBD|＝n，n＝0°，∴∠CBP＝∠PBD，····························· 4分 11

由（1）得∠ABP＝∠PBN＝∠ABN＝60°，∠PBD＝∠PBN＝30°， ············· 5分

22∴∠ABD＝∠ABP+∠PBD＝90°． ······················································· 6分

??x＋y＝180－m??x＋y＝180－m

2或?2． （2）设∠CBP＝x°，∠PBD＝y°，根据题意得：?

???x－y＝n?y－x＝n

··················································································································· 8分

180－m+2n180－m－2n

x＝x＝

44

解得： 或．

180－m－2n180－m+2ny＝y＝

44

??????

180－m+2n180－m－2n

答：情况一：∠CBP的度数是，∠PBD的度数是；

44

180－m－2n180－m+2n

情况二：∠CBP的度数是，∠PBD的度数是．

44

············································································································· 9分

27．(10分)

解：（1）垂直； ······················································································· 1分 （3）平行或垂直；（判断说理各1分，另外一种垂直和平行需简要说明，同理可得，不说扣1分） ······················································································· 6分

B

b

B

b

A

c

a

c A

a

（3）∠ACB的度数为

α+β－180360－α－βα+β

（如图1、图2）或（如图3）或（如图4）． 222

（图与结果相对应1分） ······································································ 10分

图1 图2

图3 图4

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