\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}< 1\) ;\(ĐK:x\ge0;x\ne9\)
\(\Leftrightarrow\dfrac{3\sqrt{x}}{\sqrt{x}-3}-1< 0\)
\(\Leftrightarrow\dfrac{3\sqrt{x}-\sqrt{x}+3}{\sqrt{x}-3}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+3}{\sqrt{x}-3}< 0\)
Ta có: \(2\sqrt{x}+3\ge3>0\)
\(\Rightarrow\sqrt{x}-3< 0\)
\(\Leftrightarrow\sqrt{x}< 3\)
\(\Leftrightarrow x< 9\)
Vậy \(S=\left\{x|x< 9\right\}\)