题目内容
计算:
(1)
-
;
(2)(
-
)÷
.
(1)
| x2+1 |
| x-2 |
| 3-4x |
| 2-x |
(2)(
| 1 |
| x2-2x |
| 1 |
| x2-4x+4 |
| 2 |
| x2-2x |
分析:(1)通分后相加,因式分解后约分即可;
(2)将除法转化为乘法,用分配律相乘后,化简后通分.
(2)将除法转化为乘法,用分配律相乘后,化简后通分.
解答:解:(1)原式=
-
=
=
=
=x-2;
(2)原式=
•
-
•
=
-
=
-
=
=-
.
| x2+1 |
| x-2 |
| 3-4x |
| x-2 |
=
| x2+1+3-4x |
| x-2 |
=
| x2-4x+4 |
| x-2 |
=
| (x-2)2 |
| x-2 |
=x-2;
(2)原式=
| 1 |
| x2-2x |
| x2-2x |
| 2 |
| 1 |
| (x-2)2 |
| x(x-2) |
| 2 |
=
| 1 |
| 2 |
| x |
| 2(x-2) |
=
| x-2 |
| 2(x-2) |
| x |
| 2(x-2) |
=
| x-2-x |
| 2(x-2) |
=-
| 1 |
| x-2 |
点评:本题主要考查分式的混合运算,通分、因式分解和约分是解答的关键.
练习册系列答案
相关题目