题目内容
计算:(1)(x2-y2)•
| xy |
| x2+2xy+y2 |
| x2-xy |
| x+y |
(2)
| 2x |
| x2-9 |
| 1 |
| 3-x |
| 2 |
| x2+6x+9 |
分析:(1)先进行因式分解,再约分即可;
(2)先进行因式分解,再通分,按同分母的分式进行计算即可.
(2)先进行因式分解,再通分,按同分母的分式进行计算即可.
解答:解:(1)原式=(x+y)(x-y)×
×
=y;
(2)原式=
-
-
=
-
-
=
=
.
| xy |
| (x+y)2 |
| x+y |
| x(x-y) |
=y;
(2)原式=
| 2x |
| (x+3)(x-3) |
| 1 |
| x-3 |
| 2 |
| (x+3)2 |
=
| 2x(x+3) |
| (x-3)(x+3)2 |
| (x+3)2 |
| (x-3)(x+3)2 |
| 2(x-3) |
| (x-3)(x+3)2 |
=
| 2x2+6x-x2-6x-9-2x+6 |
| (x-3)(x+3) 2 |
=
| x+1 |
| (x+3) 2 |
点评:本题考查了分式的混合运算,是基础知识要熟练掌握.
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