a: ĐKXĐ: a>0; \(a\notin\left\{1;4\right\}\)
b: \(Q=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\sqrt{a}-2}{3}\cdot\dfrac{\sqrt{a}-1}{1}\)
\(=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\)
b: Để Q>0 thì \(\sqrt{a}-2>0\)
hay a>4
c: Thay \(a=9-4\sqrt{5}\) vào Q, ta được:
\(Q=\dfrac{\sqrt{5}-2-2}{3\left(\sqrt{5}-2\right)}=\dfrac{\sqrt{5}-4}{3\left(\sqrt{5}-2\right)}=\dfrac{-3-2\sqrt{5}}{3}\)