题目内容
已知x=| 2 |
| x |
| x-1 |
| x-2 |
| x2-1 |
| x2-x-2 |
| x2+2x+1 |
分析:先把代数式利用分式的计算法则进行化简,再把已知代入求值即可.
解答:解:原式=
+
×
=
+
×
=
+
=
,
当x=
+1,
原式=
=
+1.
故填空答案:
+1.
| x |
| x-1 |
| x-2 |
| (x-1)(x+1) |
| (x+1)2 |
| x2-x-2 |
=
| x |
| x-1 |
| x-2 |
| (x-1)(x+1) |
| (x+1)2 |
| (x-2)(x+1) |
=
| x |
| x-1 |
| 1 |
| x-1 |
=
| x+1 |
| x-1 |
当x=
| 2 |
原式=
| ||
|
=
| 2 |
故填空答案:
| 2 |
点评:本题考查了分式的计算和化简.解决这类题目关键是把握好通分与约分.分式加减的本质是通分,乘除的本质是约分.同时注意在进行运算前要尽量保证每个分式最简.
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