题目内容

计算
(1)
2x
x-y
+
2y
y-x

(2)
x2
x+y
-x+y

(3)(x-
1+x
1-x
)(
2
1+x2
-1)
; 
(4)(
3
a-2
+
12
a2-4
)÷(
2
a-2
-
1
a+2
)
分析:(1)先通分,再相减,然后约分;
(2)先通分,再相减;
(3)先通分,加减后再约分;
(4)先将括号内的部分通分,加减后将除法转化为乘法,约分即可.
解答:解:(1)原式=
2x
x-y
-
2y
x-y

=
2x-2y
x-y

=
2(x-y)
x-y

=2;

(2)原式=
x2
x+y
-
x2-y2
x+y

=
x2-x2+y2
x+y

=
y2
x+y


(3)原式=(
x-x2
1-x
-
1+x
1-x
)(
2
1+x2
-
1+x2
1+x2

=
x-x2-1-x
1-x
2-1-x2
1+x2

=-
x2+1
1-x
(1-x)(1+x)
1+x2

=-1-x;

(4)原式=(
3a+6
a2-4
+
12
a2-4
)÷(
2a+4
a2-4
-
a-2
a2-4

=
3a+18
a2-4
÷
2a+4-a+2
a2-4

=
3a+18
a2-4
÷
a+6
a2-4

=3.
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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