题目内容
已知0<x<1,则
-
=( )
(x-
|
(x+
|
A、
| ||
| B、2x | ||
C、-
| ||
| D、-2x |
分析:已知0<x<1,可得出x+
>0,x-
<0,根据二次根式的性质解答.
| 1 |
| x |
| 1 |
| x |
解答:解:∵0<x<1,
∴原式=
+
=
-
=x+
-(
-x)
=2x.
故选B.
∴原式=
[x2+(
|
[x2+(
|
=
(x+
|
(x-
|
=x+
| 1 |
| x |
| 1 |
| x |
=2x.
故选B.
点评:解答此题,要弄清二次根式的性质:
=|a|.
| a2 |
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