(鄂尔多斯专版)中考数学复习方案 第一单元 数与式 课时训练(03)分式试题-人教版初中九年级全册数

课时训练(三)分式
(限时:30分钟)
|夯实基础|
1.计算(a 2b )3·b 2
a 的结果是 () A .a 5
b 5B .a 4b 5 C .ab 5D .a 5b 6
2.化简x 2
x -1+1
1-x 的结果是 ()
A .x +1
B .x -1
C .x 2
-1D .x 2+1x -1
3.[2017·某某]化简x
x -y
−
y x+y
的结果为 ()
A .1
B .x 2+y 2
x 2-y 2 C .
x -y x+y
D .x 2+y 2
4.[2019·]如果m +n =1,那么代数式2m+n m 2-mn
+1
m ·(m 2-n 2)的值为 ()
A .-3
B .-1
C .1
D .3
5.[2019·某某]计算a 2
a -1-a -1的正确结果是 () A .-1a -1
B .
1
a -1
C .-2a -1a -1
D .
2a -1a -1
6.[2019·某某]若分式
1
2x -1有意义,则x 的取值X 围是.
7.[2018·滨州]若分式x 2-9
x -3的值为0,则x 的值为. 8.[2018·某某]化简:2a
a 2-4−1
a -2=. 9.[2017·某某]若a
b =2
3,则
a+b b
=.
10.[2019·某某]化简:1-a -1
a+2÷a 2-1
a 2+4a+4
=.
11.[2019·某某]化简m m 2-4÷1+2
m -2.
12.[2019·某某]先化简,再求值:x -3
x 2+6x+9
÷1-
6
x+3
,其中x =√2-3.
13.[2019·某某]先化简,再求值:a 2-4a 2-4a+4−12-a ÷2
a 2-2a
.其中a 满足a 2+3a -2=0.
14.[2017·某某]先化简,再求值:a -6a 2-4−3a+2÷a
a -2
,其中a =20170
+-15-1
+
√27tan30°
.
15.[2018·某某]先化简,再求值:x -3
x 2-1÷x -3
x 2+2x+1-1
x -1+1,其中x 是不等式组{5x -3>3(x +1),
12x -1<9-32
x 的整数解.
|能力提升|
16.[2017·某某]若a 2-ab =0(b ≠0),则a a+b
等于 ()
A .0
B .1
2
C .0或1
2
D .1或2
17.[2018·某某]已知x +y =4√3,x -y =√3,则式子x -y +4xy x -y
x +y -
4xy
x+y
的值是 ()
A .48
B .12√3
C .16
D .12
18.[2019·滨州]观察下列一组数: a 1=1
3,a 2=3
5,a 3=6
9,a 4=10
17,a 5=15
33,…,
它们是按一定规律排列的,请利用其中规律,写出第n 个数a n =.(用含n 的式子表示) 19.[2017·某某]已知实数m 满足m 2-3m +1=0,则代数式m 2+19m 2+2
的值等于.
20.[2017·某某]若m -3
m -1·|m |=m -3
m -1,则m =. 21.若x +1
x =3,求x 2x 4+x 2+1的值.
【参考答案】
1.A
2.A
3.B
4.D[解析]2m+n m 2-mn +1
m
·(m 2-n 2
)=2m+n m (m -n )
+m -n m (m -n )·(m +n )(m -n )=3m
m (m -n )·(m +n )(m -n )=3(m +n ),∵m +n =1,∴原式=3, 故选D .
5.B[解析]原式=a 2
a -1-(a +1)=a 2
a -1−a 2-1a -1
=1
a -1.
故选B . 6.x ≠1
27.-38.
1a+2
9.53
[解析]解法1:利用比例的基本性质“两内项积等于两外项积”求解.∵a b
=23
,∴3a =2b ,a =23
b ,∴
a+b b
=
2
3
b+b b
=
53
b b =
5
3
.解法2:设参数法求解,设a =2k ,则b =3k ,∴a+b b
=
2k+3k 3k
=5k 3k =53.解法3:逆用同分母分式加减法则求解.
a+b b
=a
b +
b
b =a b +1=23+1=5
3. 10.-1
a+1
11.解:原式=m
(m+2)(m -2)÷m -2+2
m -2 =m
(m+2)(m -2)÷m
m -2 =m (m+2)(m -2)×
m -2m
=
1m+2
.
12.解:原式=
x -3
(x+3)
2
÷
x -3
x+3
=
x -3(x+3)
2
×
x+3x -3
=
1
x+3
,
当x =√2-3时,原式=√
2-3+3=√
2
=√2
2
. 13.解:a 2-4a 2-4a+4−
12-a
÷2a 2-2a =
(a -2)(a+2)(a -2)2
+
1a -2
·
a (a -2)2
=
a+2a -2
+1
a -2·
a (a -2)2
=
a+3
a -2
·a (a -2)2
=
a (a+3)2
=
a 2+3a 2
,
∵a 2+3a -2=0,∴a 2+3a =2, ∴原式=2
2=1. 14.解:原式=a -6(a+2)(a -2)−3(a -2)(a+2)(a -2)·a -2
a
=a -6-3(a -2)
(a+2)(a -2)·
a -2
a
=-2a
(a+2)(a -2)·a -2a
=-2
a+2.
∵a =20170+-15
-1
+√27tan30° =1-5+3√3×
√3
3
=-1, ∴原式=-2
-1+2=-2. 15.解:原式=
x -3(x -1)(x+1)
·
(x+1)2x -3
−(
1x -1
+
x -1
x -1
)=x+1x -1
−
x x -1
=
1x -1
.
解不等式组{5x -3>3(x +1),
12x -1<9-32x ,得3<x<5.
∵x 是整数,∴x =4.
将x =4代入得,原式=1
4-1=1
3.
16.C[解析]∵a 2-ab =0(b ≠0),∴a (a -b )=0.∴a =0或a -b =0,即a =0或a =b.∴a
a+b =0或a
a+b =1
2. 17.D[解析]x -y +4xy
x -y
x +y -4xy
x+y =
(x -y )2+4xy
x -y
·
(x+y )2-4xy
x+y
=
(x+y )2x -y
·(x -y )2
x+y =(x +y )(x -y ).
当x +y =4√3,x -y =√3时,原式=4√3×√3=12. 18.
n (n+1)2(2n +1)
[解析]这组分数的分子分别为1,3=2+1,6=3+2+1,10=4+3+2+1,15=5+4+3+2+1,…,则第n 个数的分子为
n (n+1)2
;分母分别为3=2+1,5=22+1,9=23+1,17=24+1,33=25+1,…,则第n 个数的分母是2n +1,所以第n 个数
a n =
n (n+1)
2
·
1
2n +1
=
n (n+1)
2(2n +1)
.
19.9[解析]由m 2-3m +1=0,可得m 2=3m -1.将m 2=3m -1代入m 2+19
m 2+2,得3m -1+19
3m -1+2=3m -1+19
3m+1=(3m -1)(3m+1)
3m+1
+
19
3m+1
=9m 2+183m+1
=
9(m 2+2)3m+1
=
9(3m+1)3m+1.很显然3m +1≠0,所以
9(3m+1)3m+1
=9.
20.3或-1[解析]当|m |=1时,m -3m -1
·|m |=m -3m -1
,则m =±1.因为m -1≠0,所以m =-1.当m -3=0时,m -3m -1
·|m |=
m -3m -1
,m =3,此
时m -1≠0.∴m =3或-1. 21.解:原式=1
x 2+1+1x
2
=
1
(x+1
x
) 2-1
=
132-1
=
19-1
=1
8
.。