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Đáp án:

`B=(\frac{1}{1+\sqrt{x}}-\frac{1}{1-x}):(\sqrt{x}-\frac{\sqrt{x}}{1-\sqrt{x}})` `(đk:x\ge0;x\ne1)`

`=[\frac{1.(1-\sqrt{x})}{(1-\sqrt{x}).(1+\sqrt{x})}-\frac{1}{(1-\sqrt{x}).(1+\sqrt{x})}]:[\frac{\sqrt{x}.(1-\sqrt{x})}{1-\sqrt{x}}-\frac{\sqrt{x}}{1-\sqrt{x}}]`

`=[\frac{1-\sqrt{x}-1}{(1-\sqrt{x}).(1+\sqrt{x})}]:(\frac{\sqrt{x}-x-\sqrt{x}}{1-sqrt{x}})`

`=\frac{-\sqrt{x}}{(1-\sqrt{x}).(1+\sqrt{x})}:\frac{-x}{1-\sqrt{x}}`

`=\frac{-\sqrt{x}}{(1-\sqrt{x}).(1+\sqrt{x})}.\frac{1-\sqrt{x}}{-x}`

`=\frac{\sqrt{x}.(1-\sqrt{x})}{(1-\sqrt{x}).(1+\sqrt{x}).x}`

`=\frac{\sqrt{x}}{(1+\sqrt{x}).\sqrt{x}.\sqrt{x}}`

`=\frac{1}{\sqrt{x}.(\sqrt{x}+1)}`

Vậy `B=\frac{1}{\sqrt{x}.(\sqrt{x}+1)}` với `x\ge0;x\ne1`