题目内容
已知2x=2-
|
| x | ||
|
| ||
| x |
分析:首先求出x的值,再将分式通分,然后将x的值代入即可解答.
解答:解:∵2x=
,
∴x=
,
S=
+
=
∵
•x
=
•x
=
•
=
,
∴S=
+
=4.
2-
|
∴x=
| ||||
| 2 |
S=
| x | ||
|
| ||
| x |
=
| x2+1-x2 | ||
|
∵
| 1-x2 |
=
1-
|
=
| ||||
| 2 |
| ||||
| 2 |
=
| 1 |
| 4 |
∴S=
| x | ||
|
| ||
| x |
=4.
点评:此题主要考查二次根式的性质:a=(
)2(a≥0).此题注意借助因式分解的知识达到约分化简的目的.
| a |
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