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2020-2021学年度第一学期期中考试卷参考答案
九年级数学(HS)
一、选择题(每小题3分,共30分)
1.A 2.D 3.B 4.C 5.A 6.B 7.C 8.A 9.D 10.B
二、填空题(每小题3分,共15分)
11.-1
3
12.k≥
1
2
13.- -槡 a 14.4或
9
4
15.(-8,-3)或(4,3)
三、解答题(共8题,共75分)
16.解:(1)原式=2(槡3-1)(槡3+1) 槡-3÷33-( 槡3-23+1)
=2×(3-1)-槡3
3 槡
-4+23=4+槡53
3
-4=槡53
3
. 5分
!!!!!!!!!!!!!!!!!
(2)2x2-7x-4=0,
(x-4)(2x+1)=0,
∴x-4=0或2x+1=0,
∴x1=4,x2=-
1
2
; 5分
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
(3)x2+4x+4=(3x+1)2,
(x+2)2=(3x+1)2,
(x+2)=±(3x+1),
解得:x1=
1
2
,x2=-
3
4
. 5分
!!!!!!!!!!!!!!!!!!!!!!!!!!!!
17.解:x-2
x-1
÷(x+1- 3
x-1
)
=x-2
x-1
÷[(x+1)(x-1)
x-1
- 3
x-1
]=x-2
x-1
÷x
2-4
x-1
=x-2
x-1
× x-1
(x+2)(x-2)
= 1
x+2
4分
!!!!!!!!!!!!!!!!!!!!!!!!!!!
当x 槡=3-2时,原式=
1
槡3-2+2
=槡3
3
. 6分
!!!!!!!!!!!!!!!!!!!!!!!!
18.(1)证明:∵∠BCA=∠EDA,∠BAE=∠BAE,
∴△ABC∽△ADE; 3分!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
(2)证明:∵△ABC∽△ADE,∴∠B=∠E,
∵∠DFB=∠CFE,∴△DFB∽△CFE∴
DF
CF
=FB
FE
,∴DF·EF=FC·FB. 7分
!!!!!!!!
19.(1)证明:∵m≠0,
△=(m+2)2-4m×2=m2-4m+4=(m-2)2,
而(m-2)2≥0,即△≥0,∴方程总有两个实数根; 4分!!!!!!!!!!!!!!!!!
(2)解:(x-1)(mx-2)=0,
x-1=0或mx-2=0,∴x1=1,x2=
2
m
,
当m为正整数1或2时,x2为整数,即方程的两个实数根都是整数,
∴正整数m的值为1或2. 8分
!!!!!!!!!!!!!!!!!!!!!!!!!!!
20.解:(1)如图所示,△A1B1C1即为所求;
A1(0,0),B1(3,1),C1(2,3); 3分!!!!!!!!!!!!!!!!!!!!!!!!
(2)如图所示,△AB2C2即为所求;
A2(0,-2),B2(-3,-1),C2(-2,1); 6分!!!!!!!!!!!!!!!!!!!!!!
(3)如图所示,△A3B3C3即为所求;
A3(-3,-3),B3(3,-1),C3(1,3). 9分!!!!!!!!!!!!!!!!!!!!!!!
21.解:(1)∵四边形ABCD是平行四边形,∴AB∥CD,AD∥BC,
∴∠B+∠C=180°,∠ADF=∠DEC,
∵∠AFD+∠AFE=180°,∠AFE=∠B,∴∠AFD=∠C,
∴△ADF∽△DEC; 4分!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
(2)∵AE⊥BC,AD 槡=33,AE=3,
∴在Rt△DAE中,DE= AD2+AE槡
2= 3(槡3)
2+3槡
2=6, 6分
!!!!!!!!!!!!!!
由(1)知△ADF∽△DEC,得
AE
DC
=AD
DE
,
∴AF=DC×AD
DE
= 槡4×33
6 槡
=23. 9分
!!!!!!!!!!!!!!!!!!!!!!!!!
22.解:(1)根据题意,得
(10-2x)2=81
解得x1=0.5,x2=9.5(不符合题意,舍去)
答:所剪去的小正方形的边长为0.5cm. 4分
!!!!!!!!!!!!!!!!!!!!
(2)根据题意,得
S=4x(10-2x)=-8x2+40x(0<x<5)
答:S与x的函数关系式为S=-8x2+40x,
x的取值范围为0<x<5. 7分
!!!!!!!!!!!!!!!!!!!!!!!!!!!
(3)答:不能.理由如下:-8x2+40x=60,
整理得2x2-10x+15=0
∵△=100-120=-20<0,∴此方程无解,
答:长方体盒子的侧面积为S的值不能是60cm2. 10分
!!!!!!!!!!!!!!!!!
23.解:(1)在Rt△ABC中,∵∠C=90°,AC=6,BC=8,
∴AB= 62+8槡
2=10,点