\(\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{5\sqrt{x}}{\sqrt{x}+3}=\frac{22}{x-9}\left(ĐK:x\ge0;x\ne9\right)\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)}{x-9}=\frac{22}{x-9}\)
\(\Rightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)-5\sqrt{x}\left(\sqrt{x}-3\right)=22\)
\(\Leftrightarrow x+5\sqrt{x}+6-5x+15\sqrt{x}=22\)
\(\Leftrightarrow-4x+20\sqrt{x}-16=0\)
\(\Leftrightarrow x-5\sqrt{x}+4=0\)
\(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}-4=0\\\sqrt{x}-1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=16\left(tm\right)\\x=1\left(tm\right)\end{array}\right.\)
Vậy tập nghiệm của pt đã cho là \(S=\left\{1;16\right\}\)