a) Ta có: \(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\)
\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\left(\sqrt{x}+2\right)}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
b) Để Q=2 thì \(\sqrt{x}=2\sqrt{x}-6\)
hay x=36(thỏa ĐK)
