题目内容

计算(1)
a2
a+1
-
1
a+1
(2)
a-1
a2-4a+4
×
a2-4
2a-2

(3)m+2-
4
2-m
(4)(
x
x+y
+
2y
x+y
)•
xy
x+2y
÷(
1
x
+
1
y
)
分析:(1)(3)直接通分,因式分解后约分;
(2)直接因式分解后约分;
(4)把括号里的通分,除法化为乘法,再约分.
解答:解:(1)
a2
a+1
-
1
a+1
=
a2-1
a+1
=
(a+1)(a-1)
a+1
=a-1;
(2)
a-1
a2-4a+4
×
a2-4
2a-2
=
a-1
(a-2)2
×
(a+2)(a-2)
2(a-1)
=
a+2
2(a-2)

(3)m+2-
4
2-m
=
m2-4+4
m-2
=
m2
m-2

(4)(
x
x+y
+
2y
x+y
)•
xy
x+2y
÷(
1
x
+
1
y
)
=
x+2y
x+y
xy
x+2y
xy
x+y
=
x2y2
(x+y)2
点评: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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