试题

题目:
(1)
x+y
x-y
·
1
(x+y)2

(2)
它-b
2+它b
÷
它b-2
2b2-4

(3)
x2+7x-8
4x-x3
·
x2-4
3x+24

(4)
x2+2xy+y2
xy-y2
÷
xy+y2
x2-2xy+y2

(5)x÷
1
x
·x
(6)
x3+x2
1-x2
÷x·
1
x

(7)9它2
3它
4b
·4它b2
(8)
它-4
2+4它+4
·
2+5它+6
2-它-2
÷
它+3
它-2

(9)
x2+xy
x2-xy
÷(x-y)·
x2-xy
xy

(10)(-
2
b
)2
·(
b
)3
·(
1
它b
)4

(11)[
(x-y)2
x+y
]3
·(
x2
y2-x2
)3

(12)(
x-1
x2-x-2
)2
÷
x2-2x+1
2-x
÷(
1
x2+x
)2

答案
解:(1)原式=
1
(x-y)(x+y)
=
1
x5-y5


(5)原式=
a-b
a(a+b)
×
a5(b+a)(b-a)
-a(a-b)

=a-b;

(3)原式=
(x-1)(x+8)
x(5-x)(5+x)
·
(x+5)(x-5)
3(x+8)

=
1-x
3x


(4)原式=
(x+y)5
y(x-y)
×
(x-y)5
y(x+y)

=
x5-y5
y5


(5)原式=x·x·x
=x3

(6)原式=
x5(x+1)
(1+x)(1-x)
×
1
x
·
1
x

=
1
1-x


(7)原式=9a5
4b
3a
·4ab5
=48a5b3

(8)原式=
a-4
(a+5)5
·
(a+5)(a+3)
(a+1)(a-5)
·
a-5
a+3

=
a-4
(a+5)(a+1)


(9)原式=
x(x+y)
x(x-y)
·
1
x-y
·
x(x-y)
xy

=
x+y
y(x-y)


(10)原式=
a4
b5
·
b3
a3
·
1
a4b4

=
1
a3b3


(11)原式=[
(x-y)5
x+y
·
x5
-(x+y)(x-y)
]3
=[
x-y
-(x+y)5
]3
=-
(x-y)3
(x+y)6


(15)原式=[
x-1
(x+1)(x-5)
]5·
5-x
(x-1)5
·x5(x+1)5
=
(x-1)5
(x+1)5(x-5)5
·
5-x
(x-1)5
·x5(x+1)5
=
x5
5-x

解:(1)原式=
1
(x-y)(x+y)
=
1
x5-y5


(5)原式=
a-b
a(a+b)
×
a5(b+a)(b-a)
-a(a-b)

=a-b;

(3)原式=
(x-1)(x+8)
x(5-x)(5+x)
·
(x+5)(x-5)
3(x+8)

=
1-x
3x


(4)原式=
(x+y)5
y(x-y)
×
(x-y)5
y(x+y)

=
x5-y5
y5


(5)原式=x·x·x
=x3

(6)原式=
x5(x+1)
(1+x)(1-x)
×
1
x
·
1
x

=
1
1-x


(7)原式=9a5
4b
3a
·4ab5
=48a5b3

(8)原式=
a-4
(a+5)5
·
(a+5)(a+3)
(a+1)(a-5)
·
a-5
a+3

=
a-4
(a+5)(a+1)


(9)原式=
x(x+y)
x(x-y)
·
1
x-y
·
x(x-y)
xy

=
x+y
y(x-y)


(10)原式=
a4
b5
·
b3
a3
·
1
a4b4

=
1
a3b3


(11)原式=[
(x-y)5
x+y
·
x5
-(x+y)(x-y)
]3
=[
x-y
-(x+y)5
]3
=-
(x-y)3
(x+y)6


(15)原式=[
x-1
(x+1)(x-5)
]5·
5-x
(x-1)5
·x5(x+1)5
=
(x-1)5
(x+1)5(x-5)5
·
5-x
(x-1)5
·x5(x+1)5
=
x5
5-x
考点梳理
分式的乘除法.
(1)直接根据分式的乘法法则进行计算即可;
(2)(4)直接根据分式的除法法则进行计算即可;
(3)根据分式的乘法法则进行计算即可;
(5)、(6)、(7)根据分式的乘法及除法法则进行计算即可;
(8)、(9)、(10)、(11)、(12)根据分式混合运算的法则进行计算即可.
本题考查的是分式的乘除法,熟知分式的乘法及除法法则是解答此题的关键.
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