题目内容
计算:(1)
| 12 |
| m2-9 |
| 2 |
| m-3 |
(2)
| 1 |
| x+1 |
| x+2 |
| x2-1 |
| x2-2x+1 |
| x2+4x+4 |
分析:(1)找出最小公倍数m2-9,先通分,然后合并同类项即可;
(2)先算除法,化简后再算加法.
(2)先算除法,化简后再算加法.
解答:解:(1)原式=
-
=
=
=-
;
(2)原式=
-
•
=
-
=
=
.
故答案为-
、
.
| 12 |
| (m+3)(m-3) |
| 2(m+3) |
| (m+3)(m-3) |
=
| 12-2m-6 |
| (m+3)(m-3) |
=
| -2(m-3) |
| (m+3)(m-3) |
=-
| 2 |
| m+3 |
(2)原式=
| 1 |
| x+1 |
| x+2 |
| (x+1)(x-1) |
| (x-1)2 |
| (x+2)2 |
=
| 1 |
| x+1 |
| x-1 |
| (x+1)(x+2) |
=
| x+2-x+1 |
| (x+1)(x+2) |
=
| 3 |
| x2+3x+2 |
故答案为-
| 2 |
| m+3 |
| 3 |
| x2+3x+2 |
点评:本题主要考查分式的混合运算,难度一般,熟练掌握通分、因式分解和约分的知识点.
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