题目内容
计算:
(1)(x+1)÷(2+
);
(2)(1-
)÷
.
(1)(x+1)÷(2+
| 1+x2 |
| x |
(2)(1-
| a2+8 |
| a2+4a+4 |
| 4a-4 |
| a2+2a |
考点:分式的混合运算
专题:
分析:(1)首先将分式恒等变形,然后因式分解,约分即可解决问题.
(2)先通分运算,再因式分解,约分即可解决问题.
(2)先通分运算,再因式分解,约分即可解决问题.
解答:解:(1)原式=)(x+1)÷
,
=(x+1)×
=
;
(2)原式=
×
=
×
=
.
| x2+2x+1 |
| x |
=(x+1)×
| x |
| (x+1)2 |
=
| x |
| x+1 |
(2)原式=
| a2+4a+4-a2-8 |
| (a+2)2 |
| a(a+2) |
| 4(a-1) |
=
| 4(a-1) |
| a+2 |
| a |
| 4(a-1) |
=
| a |
| a+2 |
点评:考查了分式的混合运算问题;解题的关键是准确运算,灵活因式分解.
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