题目内容
计算:(1)x+1-
| x2 |
| x+1 |
(2)先化简,再求值:
| b2-a2 |
| a2b-ab2 |
| a2+b2 |
| 2ab |
分析:(1)先通分,再计算;
(2)先通分,然后进行四则运算,最后将a=-2,b=3代入.
(2)先通分,然后进行四则运算,最后将a=-2,b=3代入.
解答:解:(1)原式=
-
=
;
(2)原式=
÷
=-
×
=-
,
∵a=-2,b=3,
∴原式=-
=-
=-2.
| (x+1)2 |
| x+1 |
| x2 |
| x+1 |
| 2x+1 |
| x+1 |
(2)原式=
| (b+a)(b-a) |
| ab(a-b) |
| 2ab+a2+b2 |
| 2ab |
=-
| a+b |
| ab |
| 2ab |
| (a+b)2 |
=-
| 2 |
| a+b |
∵a=-2,b=3,
∴原式=-
| 2 |
| a+b |
=-
| 2 |
| -2+3 |
=-2.
点评:本题考查了分式的化简求值,解答此题的关键是把分式化到最简,然后代值计算.
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