`a)` `Q=(1/{\sqrt{x}-1}-1/\sqrt{x}):({\sqrt{x}+1}/{\sqrt{x}-2}-{\sqrt{x}+2}/{\sqrt{x}-1})`
$ĐKXĐ: \begin{cases}x\ge 0\\\sqrt{x}-1\ne 0\\\sqrt{x}\ne 0 \\\sqrt{x}-2\ne 0\end{cases}$`<=>`$\begin{cases}x\ge 0\\x\ne 1\\x\ne 0 \\x\ne 4\end{cases}$`=>x>0;x\ne 1;x\ne 4`
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`b)` `Q=(1/{\sqrt{x}-1}-1/\sqrt{x}):({\sqrt{x}+1}/{\sqrt{x}-2}-{\sqrt{x}+2}/{\sqrt{x}-1})`
`={\sqrt{x}-(\sqrt{x}-1)}/{\sqrt{x}(\sqrt{x}-1)}:{(\sqrt{x}+1)(\sqrt{x}-1)-(\sqrt{x}+2)(\sqrt{x}-2)}/{(\sqrt{x}-2)(\sqrt{x}-1)}`
`=1/{\sqrt{x}(\sqrt{x}-1)}:{x-1-(x-4)}/{(\sqrt{x}-2)(\sqrt{x}-1)}`
`=1/{\sqrt{x}(\sqrt{x}-1)}: 3/{(\sqrt{x}-2)(\sqrt{x}-1)}`
`=1/{\sqrt{x}(\sqrt{x}-1)} . {(\sqrt{x}-2)(\sqrt{x}-1)}/3`
`={\sqrt{x}-2}/{3\sqrt{x}}`
Vậy `Q={\sqrt{x}-2}/{3\sqrt{x}}` với `x>0;x\ne 1;x\ne 4`
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`c)` Để `Q>0`
`<=>{\sqrt{x}-2}/{3\sqrt{x}}>0`
`<=>\sqrt{x}-2>0` (vì `3\sqrt{x}>0` với `x>0)`
`<=>\sqrt{x}-2>0`
`<=>\sqrt{x}>2`
`<=>(\sqrt{x})^2>2^2`
`<=>x>4`
Vậy `x>4` thì `Q>0`
