题目内容
先化简,再求值:(| x |
| x+y |
| 2y |
| x+y |
| xy |
| x+2y |
| 1 |
| x |
| 1 |
| y |
分析:前边括号中的分母相等,可先进行合并,后面括号式子进行通分,再把除法换算成乘法,进而化简求值.
解答:解:原式=
•
÷
=
•
=
=(
)2
=(
)2
=
.
| x+2y |
| x+y |
| xy |
| x+2y |
| x+y |
| xy |
=
| xy |
| x+y |
| xy |
| x+y |
=
| x2y2 |
| (x+y)2 |
=(
| xy |
| (x+y) |
=(
| 2 |
| 3 |
=
| 4 |
| 9 |
点评:本题主要考查分式的化简求值,有括号的先化简括号内,通过通分或因式分解等,最终化为最简,代入求值.
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