题目内容
先化简,再求值:(| 2x |
| x-3 |
| x |
| x+3 |
| x2-9 |
| x2 |
| 2 |
分析:利用乘法的分配律去括号得到原式=
•
-
•
,然后约分,再通分得到原式=
;最后把x=
代入计算即可.
| 2x |
| x-3 |
| (x+3)(x-3) |
| x2 |
| x |
| x+3 |
| (x+3)(x-3) |
| x2 |
| x+9 |
| x |
| 2 |
解答:解:原式=
•
-
•
=
-
=
;
当x=
时,原式=
=
=
.
| 2x |
| x-3 |
| (x+3)(x-3) |
| x2 |
| x |
| x+3 |
| (x+3)(x-3) |
| x2 |
=
| 2x+6 |
| x |
| x-3 |
| x |
=
| x+9 |
| x |
当x=
| 2 |
| x+9 |
| x |
| ||
|
9
| ||
| 2 |
点评:本题考查了分式的化简求值:先把分式的分子和分母因式分解,然后把括号内通分(或利用乘法的分配律去括号),再约分得到最简分式,最后把满足条件的字母的值代入计算即可.
练习册系列答案
相关题目