6) ĐK:\(x\ne9,x>0\)
a) \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9}{9-x}\right):\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right].\dfrac{x-3\sqrt{x}}{3\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{3\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}=\dfrac{-3\left(\sqrt{x}+3\right).\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)\left(3\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}}=\dfrac{-3\sqrt{x}}{3\sqrt{x}+1}-\dfrac{1}{\sqrt{x}}=\dfrac{-3\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(3\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{\sqrt{x}\left(3\sqrt{x}+1\right)}=\dfrac{-3x-3\sqrt{x}-1}{\sqrt{x}\left(3\sqrt{x}+1\right)}=\dfrac{-3x-3\sqrt{x}-1}{3x+\sqrt{x}}\)b) Để P<-1\(\Leftrightarrow\dfrac{-3x-3\sqrt{x}-1}{3x+\sqrt{x}}< -1\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}-3x-3\sqrt{x}-1< 0\\-3x-3\sqrt{x}-1< -\left(3x+\sqrt{x}\right)\end{matrix}\right.\)(vì \(3x+\sqrt{x}>0\))
\(\Leftrightarrow\left\{{}\begin{matrix}3x+3\sqrt{x}+1>0\\x>0\end{matrix}\right.\)\(\Leftrightarrow x>0\)(vì 3x+3\(\sqrt{x}\)+1>0)
Kết hợp với điều kiện
Vậy x>0 và \(x\ne9\) thì P<-1
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 \(S=\varnothing\)
