试题

题目:
根据分式的基本性质,完成下列各等式.
(1)
1
a
=
(     )
ab
(b≠0)

(2)
xy2
x2y
=
(    )
x

(3)
3a
a+b
=
6ab
(     )
(b≠0);
(4)
x2y
xy2
=
(    )
y
(x≠0且y≠0)

(5)3x-2=
(     )
3x+2
(x≠
2
3
)

(6)
2m
m-n
=
(         )
(n-m)2

答案
解:(1)
1
a
=
b
ab
(b≠0);
(2)
xy2
x2y
=
y
x

(3)
3a
a+b
=
6ab
2ab+2b2
(b≠0;
(4)
x2y
xy2
=
x
y
,(x≠0且y≠0);
(5)3x-2=
9x2-4
3x+2
(x≠
2
3
)

(6)
2m
m-n
=
2m2-2mn
(n-m)2

故答案为;b,y,2ab+2b2,x,9x2-4,2m2-2mn;
解:(1)
1
a
=
b
ab
(b≠0);
(2)
xy2
x2y
=
y
x

(3)
3a
a+b
=
6ab
2ab+2b2
(b≠0;
(4)
x2y
xy2
=
x
y
,(x≠0且y≠0);
(5)3x-2=
9x2-4
3x+2
(x≠
2
3
)

(6)
2m
m-n
=
2m2-2mn
(n-m)2

故答案为;b,y,2ab+2b2,x,9x2-4,2m2-2mn;
考点梳理
分式的基本性质.
根据分式的基本性质,分式的分子与分母同乘(或除以)一个不等于0的整式,分式的值不,从而求出答案.
此题考查了分式的基本性质,一定要熟练掌握分式的基本性质是解题的关键,是一道基础题.
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