题目内容
计算:
(1)
-
(2)(1+
)÷(2x-
).
(1)
| 1 |
| x-1 |
| x |
| x2-1 |
(2)(1+
| 1 |
| x |
| 1+x2 |
| x |
分析:(1)先确定最简公分母(x+1)(x-1),通分得到原式=
-
,然后进行同分母的分式的减法运算;
(2)先把括号内进行通分得到原式=
÷
,再把除法运算化为乘法运算和分母分解因式,然后约分即可.
| x+1 |
| (x+1)(x-1) |
| x |
| (x+1)(x-1) |
(2)先把括号内进行通分得到原式=
| x+1 |
| x |
| 2x2-(1+x2) |
| x |
解答:解:(1)原式=
-
=
=
;
(2)原式=
÷
=
•
=
.
| x+1 |
| (x+1)(x-1) |
| x |
| (x+1)(x-1) |
=
| x+1-x |
| (x+1)(x-1) |
=
| 1 |
| x2-1 |
(2)原式=
| x+1 |
| x |
| 2x2-(1+x2) |
| x |
=
| x+1 |
| x |
| x |
| (x+1)(x-1) |
=
| 1 |
| x-1 |
点评:本题考查了分式的混合运算:先把分式的分子或分母因式分解(有括号,先算括号),再进行约分,然后进行分式的加减运算.
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