题目内容
计算:[| m-n |
| m2-2mn+n2 |
| mn+n2 |
| m2-n2 |
| mn |
| n-1 |
分析:先对m2-2mn+n2、mn+n2、m2-n2分解因式,再通分,最后化简.
解答:解:原式=[
-
]•
=(
-
)•
=
•
=-
=
.
故答案为
.
| m-n |
| (m-n)2 |
| n(m+n) |
| (m+n)(m-n) |
| mn |
| n-1 |
=(
| 1 |
| m-n |
| n |
| m-n |
| mn |
| n-1 |
=
| 1-n |
| m-n |
| mn |
| n-1 |
=-
| mn |
| m-n |
=
| mn |
| n-m |
故答案为
| mn |
| n-m |
点评:主要考查分式的化简.注意去括号时不要漏项.
练习册系列答案
相关题目