Ta viết lại đẳng thức: \(P\left(\sqrt{x}-3\right)=2\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}=\dfrac{3P-1}{P-2}\ge1\)
\(\Rightarrow\dfrac{3P-1}{P-2}-1\ge0\Leftrightarrow\dfrac{2P+1}{P-2}\ge0\)
Trường hợp 1:
\(\left\{{}\begin{matrix}2P+1\ge0\\P-2>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P\ge-\dfrac{1}{2}\\P>2\end{matrix}\right.\Rightarrow P>2\)
Trường hợp 2: \(\left\{{}\begin{matrix}2P+1\le0\\P-2< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}P\le-\dfrac{1}{2}\\P< 2\end{matrix}\right.\Rightarrow P\le-\dfrac{1}{2}\).
Vậy: \(P_{max}=-\dfrac{1}{2}\Rightarrow\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}=-\dfrac{1}{2}\)
\(\Rightarrow x=1.\)
