七年级数学期末模拟卷(参考答案)【江苏南京】七年级数学期末模拟卷(江苏南京专用)-学易金卷:2023-2024学年初中上学期期末模拟考试.docx

七年级数学期末模拟卷(参考答案)【江苏南京】七年级数学期末模拟卷(江苏南京专用)-学易金卷:2023-2024学年初中上学期期末模拟考试.docx

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2023-2024学年江苏南京上学期期末模拟考试

七年级数学

1

2

3

4

5

6

7

8

D

C

A

D

A

B

B

C

9.5.5×104 10.6 11.36

12.1 13.﹣4 14.4

15.±1或﹣3 16.x-614=x-6-4.55

18.198

19.(8分)

解:(1)原式=3

=(31

=9+5

=14;··············································································4分

(2)原式=

=-

=﹣8.·············································································8分

20.(6分)

解:(1)去括号得:4x+1=6﹣2x,

移项合并得:6x=5,

解得:x=56;

(2)解:去分母,得:2(2x+1)﹣(x﹣1)=6,

去括号,得:4x+2﹣x+1=6,

移项,得:4x﹣x=6﹣2﹣1,

合并同类项,得:3x=3,

系数化为1,得:x=1.·······························································6分

21.(6分)

解:原式=﹣x2﹣y+4x+2x2﹣4x﹣2y

=x2﹣3y,··········································································3分

当x=﹣3,y=﹣1时,

原式=(﹣3)2﹣3×(﹣1)=9+3=12.···············································6分

22.(8分)

解:(1)如图所示:

······························6分

(2)在这个几何体上再添加一些小正方体,并保持俯视图和左视图不变,最多可以再添加3个小正方体,

故答案为:3.·········································································8分

23.(6分)

解:(1)设商场第一次购进x套运动服,

由题意得:680002x

解这个方程,得x=200.

经检验,x=200是所列方程的根.

2x+x=2×200+200=600.

答:商场两次共购进这种运动服600套.·················································3分

(2)第一批运动服的进价

第二批运动服的进价

设每套运动服的售价是x元,

由题意得:400(x﹣170)﹣200(x﹣160)=12000,

解得:x=240

故答案为:240.······································································6分

24.(6分)

解:如图所示,

(1)线段AD即为所求;·······························································2分

(2)AC=2

∵AC2=20,CD2=5,AD2=25,

∴AC2+CD2=AD2

∴△ACD为直角三角形;

故答案为:25、直角三角形.···························································4分

(3)∵AD∥BC,

∴∠BAC=∠ACD=90°,

∵E为BC的中点,

∴AE=12BC=1

故答案为:52.·········································································6

25.(6分)

解:(1)如图1所示:

∵AC=AB+BC,AB=6cm,BC=4cm

∴AC=6+4=10cm

又∵D为线段AC的中点

∴DC=12AC=12

∴DB=DC﹣BC=6﹣5=1cm·······················································2分

(2)如图2所示:

设BD=xcm

∵BD=14AB

∴AB=4BD=4xcm,CD=3BD=3xcm,

又∵DC=DB+BC,

∴BC=3x﹣x=2x,

又∵AC=AB+BC,

∴AC=4x+2x=6xcm,

∵E为线段AB的中点

∴BE=12AB=12×

又∵EC=BE+BC,

∴EC=2x+2x=4

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