题目内容
先化简(x |
x-1 |
1 |
x+1 |
1 |
x2-1 |
分析:本题需先把括号中的每一项分别进行相乘,再把所得结果进行相加,再把x的值代入即可求出结果.
解答:解:原式=
-
,
=
,
=
,
∴(
-
)÷
=
•
=x2+1.
取x=0代入上式得,
=02+1
=1.
x(x+1) |
(x-1)(x+1) |
x-1 |
(x-1)(x+1) |
=
x2+x-x+1 |
(x-1)(x+1) |
=
x2+1 |
x2-1 |
∴(
x |
x-1 |
1 |
x+1 |
1 |
x2-1 |
x2+1 |
x2-1 |
x2-1 |
1 |
取x=0代入上式得,
=02+1
=1.
点评:本题主要考查了分式的化简求值,在解题时要注意分式的运算顺序和法则是解题的关键.
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