题目内容
先化简(1+1 |
x |
x2-1 |
x2 |
分析:先把括号内通分,再把x2-1分解为(x+1)(x-1),然后约分,要使原分式有意义,x只能取2,最后代入计算即可.
解答:解:(1+
)÷
,
=
•
,
=
.
当x是满足-2<x≤2的整数,x=-1,0,1,2.
而要使原分式有意义,x≠±1,且x≠0,
即x=2,
当x=2时,原式=
=2.
1 |
x |
x2-1 |
x2 |
=
x+1 |
x |
x2 |
(x+1)(x-1) |
=
x |
x-1 |
当x是满足-2<x≤2的整数,x=-1,0,1,2.
而要使原分式有意义,x≠±1,且x≠0,
即x=2,
当x=2时,原式=
2 |
2-1 |
点评:本题考查了分式的化简求值:先对括号内的分式通分,去括号,再对所有分式的分子和分母进行因式分解,然后进行约分,最后代值计算;也考查了能使分式有意义的条件即各分母不为0.
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