问题
解答题
（1）x-
（2）
|
答案
（1）原式=（x-y）-（
| x2-y2 |
| x |
| x2+y2 |
| x |
=（x-y）-
| -2y2 |
| x |
=
| x-y |
| x |
| -2y2 |
| x |
=
| x2-xy+2y2 |
| x |
（2）原式=
| ||||
|
=
| ||
|
=
| 2(x+y) |
| xy |
| xy |
| x+y |
=2．
故答案为
| x2-xy+2y2 |
| x |
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（1）x-
（2）
|
（1）原式=（x-y）-（
| x2-y2 |
| x |
| x2+y2 |
| x |
=（x-y）-
| -2y2 |
| x |
=
| x-y |
| x |
| -2y2 |
| x |
=
| x2-xy+2y2 |
| x |
（2）原式=
| ||||
|
=
| ||
|
=
| 2(x+y) |
| xy |
| xy |
| x+y |
=2．
故答案为
| x2-xy+2y2 |
| x |