ĐK: \(x\ge0;x\ne4;x\ne9\)
\(Q=\left(\dfrac{x-3\sqrt{x}}{9-x}+1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}+\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(=\left[-\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+1\right]:\left[\dfrac{9-x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{\left(2-\sqrt{x}\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\right]\)
\(=\dfrac{3}{\sqrt{x}+3}:\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}.\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}-2}\)
\(=\dfrac{3}{2-\sqrt{x}}\)
help em cau 3 vs
