Với a > 0 ; a khác 1 ; a khác 4 ; a khác 5/2
\(Q=\left(\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{a-1+a-4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{2a-5}=\dfrac{\sqrt{a}-2}{\sqrt{a}\left(2a-5\right)}\)
b, Ta có Q > 0
\(\dfrac{\sqrt{a}-2}{\sqrt{a}\left(2a-5\right)}>0\Rightarrow\left\{{}\begin{matrix}a>4\\a>0\\a>\dfrac{5}{2}\end{matrix}\right.\)
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