đkxđ:\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\\\sqrt{x}\left(3-\sqrt{x}\right)\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne4\\x\ne9\end{matrix}\right.\)
a) \(Q=\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\frac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\frac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\frac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-2\right)}.\frac{\left(\sqrt{x}+2\right)}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}-3}\)
b) Để Q=2 thì \(2(\sqrt{x}-3)=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=8\)
\(\Leftrightarrow x=64\)
c) Vì \(\sqrt{x}+2\ge2>0\) nên để Q<0 thì \(\sqrt{x}< 3\) hay x<9
