题目内容
先化简,再求值:(1 |
x-y |
1 |
x+y |
2y |
x2+2xy+y2 |
3 |
2 |
3 |
2 |
分析:先对
-
通分,再对x2+2xy+y2分解因式,进行化简求值.
1 |
x-y |
1 |
x+y |
解答:解:(
-
)÷
=
•
=
•
=
,
把x=
+
,y=
-
代入上式,得
原式=
=
=
.
1 |
x-y |
1 |
x+y |
2y |
x2+2xy+y2 |
(x+y)-(x-y) |
(x-y)(x+y) |
(x+y)2 |
2y |
=
2y |
(x-y)(x+y) |
(x+y)2 |
2y |
x+y |
x-y |
把x=
3 |
2 |
3 |
2 |
原式=
(
| ||||||||
(
|
2
| ||
2
|
| ||
2 |
点评:考查分式的化简求值,注意先化简,再代值计算.
练习册系列答案
相关题目