a: Ta có: \(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\)
\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}-x}\)
\(=\dfrac{-x+2\sqrt{x}}{\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{-x+3\sqrt{x}}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}{-\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b: Để Q=2 thì \(\sqrt{x}+2=2\sqrt{x}-6\)
\(\Leftrightarrow2\sqrt{x}-\sqrt{x}=2+6=8\)
hay x=64
