题目内容
已知x-2y=0,求(| x |
| y |
| y |
| x |
| xy |
| x2-2xy+y2 |
分析:由x-2y=0,可求x=2y,然后再通分、完全平方公式对所求式子化简,再把x=2y代入求值即可.
解答:解:原式=(
-
)•
=
•
=
•
=
,
∵x-2y=0,
∴x=2y,
∴原式=
=
=
=3.
| x |
| y |
| y |
| x |
| xy |
| x2-2xy+y2 |
=
| x2-y2 |
| xy |
| xy |
| x2-2xy+y2 |
=
| (x-y)(x+y) |
| xy |
| xy |
| (x-y)2 |
=
| x+y |
| x-y |
∵x-2y=0,
∴x=2y,
∴原式=
| x+y |
| x-y |
| 2y+y |
| 2y-y |
| 3y |
| y |
点评:本题利用了完全平方公式、平方差公式、分式的加减乘除运算的知识.
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