$Q=(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}):(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1})$
`ĐKXĐ:a≥0`
`\sqrt{a}-1\ne0⇔a\ne1`
`\sqrt{a}-2\ne0⇔a\ne4`
`⇔ĐKXĐ:` `a≥0, a\ne1, a\ne4`
$Q=(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}):(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1})$
$=(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}(\sqrt{a}-1)}):(\dfrac{(\sqrt{a}-1)(\sqrt{a}+1)-(\sqrt{a}-2)(\sqrt{a}+2)}{(\sqrt{a}-1)(\sqrt{a}-2)}$
$=\dfrac{1}{\sqrt{a}(\sqrt{a}-1)}:(\dfrac{a-1-a+4}{(\sqrt{a}-1)(\sqrt{a}-2)})$
$=\dfrac{1}{\sqrt{a}(\sqrt{a}-1)}.\dfrac{(\sqrt{a}-1)(\sqrt{a}-2)}{3}$
$=\dfrac{\sqrt{a}-2}{3\sqrt{a}}$
$b)Q>0$
`->`$\dfrac{\sqrt{a}-2}{3\sqrt{a}}>0$
Ta thấy `\sqrt{a}≥0⇔3\sqrt{a}≥0∀x`
`->\sqrt{a}-2>0`
`⇔\sqrt{a}>2`
`⇔a>4`
Vậy để `Q` dương thì `a>4`
