Bài 1:
\(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(đk:x>0,x\ne9\right)\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=-\dfrac{x+3\sqrt{x}}{\left(3+\sqrt{x}\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}\)
\(=-\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3+\sqrt{x}\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\sqrt{x}+5}=\dfrac{x}{\sqrt{x}-5}\)
Bài 2:
a) \(\dfrac{1}{\sqrt{x}-2}\ge0\Leftrightarrow\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>2\)
\(\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -4\end{matrix}\right.\)
b) \(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-2< 0\\\sqrt{x}+1>0\end{matrix}\right.\)
\(\Leftrightarrow-1< \sqrt{x}< 2\)
\(\Leftrightarrow0\le x< 4\)
\(1,\\ \left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(x>0;x\ne9\right)\\ =\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{3\sqrt{x}+x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}\\ =\dfrac{-\sqrt{x}}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{5-\sqrt{x}}=\dfrac{x}{\sqrt{x}-5}\)
\(2,\\ a,\dfrac{1}{\sqrt{x}-2}\ge0\Leftrightarrow\sqrt{x}-2>0\left(\sqrt{x}-2\ne0;1>0\right)\Leftrightarrow x>4\\ b,\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\Leftrightarrow\sqrt{x}-2< 0\left(\sqrt{x}+1>0\right)\Leftrightarrow x< 4\\ c,\dfrac{2\sqrt{x}-3}{x-3\sqrt{x}+4}>0\Leftrightarrow2\sqrt{x}-3>0\left(x-3\sqrt{x}+4=\left(\sqrt{x}-\dfrac{3}{2}\right)^2+\dfrac{7}{4}>0\right)\\ \Leftrightarrow x>\dfrac{9}{4}\)
