题目内容
计算
(1)
+
+
(2)
-x+y.
(1)
| 12 |
| m2-9 |
| 2 |
| 3-m |
| 2 |
| m+3 |
(2)
| x2 |
| x+y |
分析:(1)首先通分,然后利用同分母的分式相加减的运算法则求解即可求得答案;
(2)首先将原式化为
-(x-y),再通分,然后利用同分母的分式相加减的运算法则求解即可求得答案.
(2)首先将原式化为
| x2 |
| x+y |
解答:解:(1)
+
+
=
-
+
=
=0;
(2)
-x+y
=
-(x-y)
=
-
=
=
.
| 12 |
| m2-9 |
| 2 |
| 3-m |
| 2 |
| m+3 |
=
| 12 |
| (m+3)(m-3) |
| 2(m+3) |
| (m+3)(m-3) |
| 2(m-3) |
| (m+3)(m-3) |
=
| 12-2(m+3)+2(m-3) |
| (m+3)(m-3) |
=0;
(2)
| x2 |
| x+y |
=
| x2 |
| x+y |
=
| x2 |
| x+y |
| (x+y)(x-y) |
| x+y |
=
| x2-(x+y)(x-y) |
| x+y |
=
| y2 |
| x+y |
点评:此题考查了分式的加减运算法则.此题比较简单,注意先通分,再求解,注意运算结果需化为最简.
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