Điều kiện : \(x>0;x\ne9\)
a) Với mọi \(x>0;x\ne9\), ta có :
\(C=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right]:\left[\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{1}{\sqrt{x}}\right]\)
\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{2\sqrt{x}+4}{-\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(3-\sqrt{x}\right)}{2\sqrt{x}+4}\)
\(=-\dfrac{3\sqrt{x}}{2\sqrt{x}+4}\)
b) Để \(C< -1\) thì \(-\dfrac{3\sqrt{x}}{2\sqrt{x}+4}< -1\Leftrightarrow\dfrac{3\sqrt{x}}{2\sqrt{x}+4}-1>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}-4}{2\sqrt{x}+4}>0\Rightarrow\sqrt{x}-4>0\Leftrightarrow\sqrt{x}>4\Leftrightarrow x>16\).
Kết hợp với điều kiện : Vậy \(C< -1\Leftrightarrow x>16\)
