a) ĐKXĐ: \(\left\{{}\begin{matrix}\sqrt{a}\ge0\\\sqrt{a}-1\ne0\\\sqrt{a}-2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a\ge0\\a\ne1\\a\ne4\end{matrix}\right.\)
\(Q=\left(\frac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\\ =\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\\ =\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b) Vì \(3\sqrt{a}\ge0\forall a\ge0\)nên để Q dương thì
\(\left\{{}\begin{matrix}3\sqrt{a}>0\\\sqrt{a}-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a>0\\a>4\end{matrix}\right.\Leftrightarrow a>4\)
c) Ta có: \(\sqrt{a}=\sqrt{9-4\sqrt{5}}=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)
\(Q=\frac{\sqrt{a}-2}{3\sqrt{a}-3}=\frac{\sqrt{5}-2-2}{3\left(\sqrt{5}-2\right)}=\frac{\sqrt{5}-4}{3\left(\sqrt{5}-2\right)}=\frac{-3-2\sqrt{5}}{3}\)
(cái cuối nhân lượng liên hợp để ra kết quả cuối cùng)
