题目内容
计算:
(1)
-x-1;
(2)
-
.
(1)
| x2 |
| x-1 |
(2)
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
考点:分式的加减法
专题:
分析:(1)先将整式-x-1变形为分母为x-1的分式,再根据同分母分式加减法法则计算即可;
(2)先通分,然后进行同分母分式加减运算,最后要注意将结果化为最简分式.
(2)先通分,然后进行同分母分式加减运算,最后要注意将结果化为最简分式.
解答:解:(1)
-x-1=
-
=
;
(2)
-
=
-
=
=
.
| x2 |
| x-1 |
| x2 |
| x-1 |
| x2-1 |
| x-1 |
| 1 |
| x-1 |
(2)
| x+2 |
| x2-2x |
| x-1 |
| x2-4x+4 |
| (x+2)(x-2) |
| x(x-2)2 |
| x(x-1) |
| x(x-2)2 |
| x2-4-x2+x |
| x(x-2)2 |
| x-4 |
| x3-4x2+4x |
点评:本题考查了分式的加减法,注意:分式的加减运算中,如果是同分母分式,那么分母不变,把分子直接相加减即可;如果是异分母分式,则必须先通分,把异分母分式化为同分母分式,然后再相加减.
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