题目内容
| 1 |
| a+1 |
| a+3 |
| a2-1 |
| a2-2a+1 |
| (a+1)(a-3) |
分析:首先把分式分子分母能分解因式的先分解因式,进行约分化简,然后进行减法运算,最后整体法代值计算.
解答:解:
-
×
=
-
×
=
-
=
=
=
;
当a2+2a-1=0即a2+2a=1时,原式=
=1.
| 1 |
| a+1 |
| a+3 |
| a2-1 |
| a2-2a+1 |
| (a+1)(a+3) |
=
| 1 |
| a+1 |
| a+3 |
| (a+1)(a-1) |
| (a-1)2 |
| (a+1)(a+3) |
=
| 1 |
| a+1 |
| a-1 |
| (a+1)2 |
| a+1-a+1 |
| (a+1)2 |
| 2 |
| (a+1)2 |
| 2 |
| a2+2a+1 |
当a2+2a-1=0即a2+2a=1时,原式=
| 2 |
| 1+1 |
点评:本题的关键是化简,然后把给定的值代入求值.本题中a的值无需求出,可整体代入求解.
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