ĐKXĐ: \(a>0;a\ne1;a\ne4\)
\(Q=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)=\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3\sqrt{a}\left(\sqrt{a}-1\right)}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
Để \(Q>0\Rightarrow\frac{\sqrt{a}-2}{3\sqrt{a}}>0\Rightarrow\sqrt{a}-2>0\Rightarrow\sqrt{a}>2\Rightarrow a>4\)
\(a=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\Rightarrow\sqrt{a}=\sqrt{5}-2\)
\(\Rightarrow Q=\frac{\sqrt{5}-2-2}{3\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-4}{3\sqrt{5}-6}=\frac{-3-2\sqrt{5}}{3}\)