6) \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9}{9-x}\right):\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\)
\(P=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(\sqrt{x}+3\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{3\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}\)
\(P=\dfrac{3\sqrt{x}-x+x+9}{\left(\sqrt{x}+3\right)}.\dfrac{\left(-\sqrt{x}\right)}{3\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}\)
\(P=\dfrac{3\sqrt{x}+9}{\left(\sqrt{x}+3\right)}.\dfrac{\left(-\sqrt{x}\right)}{3\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}\)
\(P=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)}.\dfrac{\left(-\sqrt{x}\right)}{3\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}\)
\(P=\dfrac{-3\sqrt{x}}{3\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}=\dfrac{-3x-\left(3\sqrt{x}+1\right)}{\sqrt{x}\left(3\sqrt{x}+1\right)}\)
\(P=\dfrac{-3x-3\sqrt{x}-1}{\sqrt{x}\left(3\sqrt{x}+1\right)}\)
\(P< -1\Leftrightarrow\dfrac{-3x-3\sqrt{x}-1}{3x+\sqrt{x}}< -1\)
\(\Leftrightarrow\dfrac{-3x-3\sqrt{x}-1}{3x+\sqrt{x}}+1< 0\)
\(\Leftrightarrow\dfrac{-3x-3\sqrt{x}-1+3x+\sqrt{x}}{3x+\sqrt{x}}< 0\)
\(\Leftrightarrow\dfrac{-2\sqrt{x}-1}{3x+\sqrt{x}}< 0\)
ta có \(3x+\sqrt{x}>0\)
\(\Leftrightarrow-2\sqrt{x}-1< 0\)
\(\Leftrightarrow-2\sqrt{x}< 1\)
\(\Leftrightarrow\sqrt{x}< \dfrac{-1}{2}\) ( vô lí )
vậy \(S=\varnothing\)