a) \(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\)
ĐKXĐ:\(x>4\)
\(Q=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x+2}\right)-2\sqrt{x}\left(\sqrt{x}-1\right)-\left(5\sqrt{x}+2\right)}{x-4}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\)
\(=\dfrac{2\sqrt{x}-x}{x-4}:\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(\sqrt{x}+2\right)^2}\) \(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{2-\sqrt{x}}{3-\sqrt{x}}\)
b) \(Q=2\Leftrightarrow\dfrac{2-\sqrt{x}}{3-\sqrt{x}}=2\) \(\Leftrightarrow\dfrac{2-\sqrt{x}}{3-\sqrt{x}}-2=0\) \(\Leftrightarrow\dfrac{\sqrt{x}-4}{3-\sqrt{x}}=0\)(*)
mà \(3-\sqrt{x}\ne0\) để biểu thức có nghĩa
(*)\(\Leftrightarrow\sqrt{x}-4=0\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\)
Vậy khi \(x=16\) thì \(Q=2\)
a) Ta có: \(Q=\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{5\sqrt{x}+2}{4-x}\right):\dfrac{3\sqrt{x}-x}{x+4\sqrt{x}+4}\)
\(=\dfrac{x+3\sqrt{x}+2-2x+4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-x+2\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\)
