试题

题目:
计算:
(1)
x2
y-x
-
y2
y-x
;                  (2)
a-b
a
÷(a-
2ab-b2
a
)

答案
解:(1)原式=
x2-y2
y-x
=
(x+y)(x-y)
y-x
=-(x+y)=-x-y;
(2)原式=
a-b
a
÷
a2-2ab+b2
a
=
a-b
a
·
a
(a-b)2
=
1
a-b

解:(1)原式=
x2-y2
y-x
=
(x+y)(x-y)
y-x
=-(x+y)=-x-y;
(2)原式=
a-b
a
÷
a2-2ab+b2
a
=
a-b
a
·
a
(a-b)2
=
1
a-b
考点梳理
分式的混合运算.
(1)利用同分母的分时的减法法则,分母不变,分子相减,然后进行约分即可;
(2)首先计算括号内的式子,然后把除法转化成乘法运算,进行乘法运算即可求解.
本题主要考查分式的混合运算,通分、因式分解和约分是解答的关键.
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