a) \(ĐKXĐ:\hept{\begin{cases}x>0\\x\ne9\end{cases}}\)
\(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(\Leftrightarrow C=\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{9-x}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3\sqrt{x}+9}{9-x}:\frac{2\sqrt{x}+4}{x-3\sqrt{x}}\)
\(\Leftrightarrow C=\frac{3}{3-\sqrt{x}}\cdot\frac{x-3\sqrt{x}}{2\sqrt{x}+4}\)
\(\Leftrightarrow C=\frac{-3}{2\sqrt{x}+4}\)
b) Để \(-\frac{3}{2\sqrt{x}+4}< -1\)
\(\Leftrightarrow\frac{1+2\sqrt{x}}{2\sqrt{x}+4}< 0\)
Vì \(\hept{\begin{cases}1+2\sqrt{x}>0\\2\sqrt{x}+4>0\end{cases}\Leftrightarrow C>0}\)
Vậy để C <-1 <=> \(x\in\varnothing\)
c) \(A=\frac{1}{\sqrt{3}-\sqrt{2}}=\sqrt{3}+\sqrt{2}\)
\(\Leftrightarrow A^2=3+2+2\sqrt{5}=5+2\sqrt{5}\)
\(B=\sqrt{5}+1\)
\(\Leftrightarrow B^2=5+1+2\sqrt{5}=6+2\sqrt{5}\)
Vì \(5+2\sqrt{5}< 6+2\sqrt{5}\)
\(\Leftrightarrow A^2< B^2\)
\(\Leftrightarrow A< B\)
Vậy \(\frac{1}{\sqrt{3}-\sqrt{2}}< \sqrt{5}+1\)
