a) \(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\left(đk:x\ge0,x\ne4\right)\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)-2\sqrt{x}\left(\sqrt{x}-2\right)-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{3\sqrt{x}-x}\)
\(=\dfrac{\left(x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(3\sqrt{x}-x\right)}\)
\(=\dfrac{\left(-x+2\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(3\sqrt{x}-x\right)}=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{-\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b) \(Q=2\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=2\Leftrightarrow2\sqrt{x}-6=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=8\Leftrightarrow x=64\)
c) \(Q=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+2>0\\\sqrt{x}-3< 0\end{matrix}\right.\)(do \(\sqrt{x}+2>\sqrt{x}-3\))
\(\Leftrightarrow-2< \sqrt{x}< 3\)
\(\Leftrightarrow0\le x< 9\) và \(x\ne4\)
a: Ta có: \(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\)
\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{-\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+2}{-\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
b: Để Q=2 thì \(\sqrt{x}+2=2\sqrt{x}-6\)
\(\Leftrightarrow\sqrt{x}=8\)
hay x=64
các bạn giúp mình đc ko ạ 
