题目内容
先计算化简再求值:(1)(1+
| 1 |
| x |
| x2-1 |
| x |
(2)
| 12 |
| a2-9 |
| 2 |
| a-3 |
分析:(1)先把括号内通分计算,把x2-1分解因式,再做除法运算,约分化简,最后代值计算.
(2)先通分,再做减法化简,最后代值计算.
(2)先通分,再做减法化简,最后代值计算.
解答:解:(1)(1+
)÷
=
÷
=
×
=
.
当x=3时,原式=
=
;
(2)
-
=
-
=
=
=-
.
当a=5时,原式=-
=-
.
| 1 |
| x |
| x2-1 |
| x |
=
| x+1 |
| x |
| (x+1)(x-1) |
| x |
=
| x+1 |
| x |
| x |
| (x+1)(x-1) |
=
| 1 |
| x-1 |
当x=3时,原式=
| 1 |
| 3-1 |
| 1 |
| 2 |
(2)
| 12 |
| a2-9 |
| 2 |
| a-3 |
=
| 12 |
| (a+3)(a-3) |
| 2(a+3) |
| (a+3)(a-3) |
=
| 12-2a-6 |
| (a+3)(a-3) |
=
| -2(a-3) |
| (a+3)(a-3) |
=-
| 2 |
| a+3 |
当a=5时,原式=-
| 2 |
| 5+3 |
| 1 |
| 4 |
点评:解答此题的关键是把分式化到最简,然后代值计算.
练习册系列答案
相关题目