Lời giải :
`A=(\sqrt{x}/{\sqrt{x}+1}-x/{x-1}):({2x}/{x-1}-\sqrt{x}/{\sqrt{x}-1})`
`ĐK: x>=0;x ne 1`
`a)A=({\sqrt{x}(\sqrt{x}-1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}-x/{(\sqrt{x}+1)(\sqrt{x}-1)}):({2x}/{(\sqrt{x}+1)(\sqrt{x}-1)}-{\sqrt{x}(\sqrt{x}+1)}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`A={x-\sqrt{x}-x}/{(\sqrt{x}+1)(\sqrt{x}-1)}:{2x-x-\sqrt{x}}/{(\sqrt{x}+1)(\sqrt{x}-1)}`
`A=-\sqrt{x}/{(\sqrt{x}+1)(\sqrt{x}-1)}.{(\sqrt{x}+1)(\sqrt{x}-1)}/{x-\sqrt{x}}`
`A=-\sqrt{x}/{\sqrt{x}(\sqrt{x}-1)}`
`A=-1/{\sqrt{x}-1}`
Vậy `A=-1/{\sqrt{x}-1}` với `x>1`
`b)` Với `x>=0;x ne 1` để `A=2<=>-1/{\sqrt{x}-1}=2`
`<=>-1=2(\sqrt{x}-1)`
`<=>-1=2\sqrt{x}-1`
`<=>2\sqrt{x}=0`
`<=>x=0(tm)`
Vậy `x=0` để `A=2`
`c)` Với `x>=0;x ne 1` để `\sqrt{A^2}=A`
`<=>|A|=A`
`<=>|-1/{\sqrt{x}-1}|=-1/{\sqrt{x}-1}`
`<=>`\(\left[ \begin{array}{l}\frac{-1}{\sqrt{x}-1}=\frac{-1}{\sqrt{x}-1}(1)\\\frac{-1}{\sqrt{x}-1}=\frac{1}{\sqrt{x}-1}(2)\end{array} \right.\)
`(1)<=>\sqrt{x}-1=\sqrt{x}-1` (luôn đúng `forall x>=0; x ne 1)`
`(2)<=>-\sqrt{x}+1=\sqrt{x}-1`
`<=>-2\sqrt{x}=-2`
`<=>\sqrt{x}=1`
`<=>x=1(ktm)`
