Đặt \(P\sqrt{x}=Q\)
\(Q=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(Q=\frac{x+\sqrt{x}}{\sqrt{x}-1}\)
\(\Leftrightarrow Q\left(\sqrt{x}-1\right)=x+\sqrt{x}\)
\(\Leftrightarrow Q\sqrt{x}-Q=x+\sqrt{x}\)
\(\Leftrightarrow x+\sqrt{x}-Q\sqrt{x}+Q=0\)
\(\Leftrightarrow x-\sqrt{x}\left(Q-1\right)+Q=0\)
Phương trình trên có nghiệm khi \(\Delta\ge0\)
\(\Leftrightarrow\left(Q-1\right)^2-4Q\ge0\)
\(\Leftrightarrow Q^2-2Q+1-4Q\ge0\)
\(\Leftrightarrow Q^2-6Q+1\ge0\)
\(\Leftrightarrow Q^2-6Q+9-8\ge0\)
\(\Leftrightarrow\left(Q-3\right)^2\ge8\)
\(\Leftrightarrow\left[{}\begin{matrix}Q-3\ge\sqrt{8}\\Q-3\le-\sqrt{8}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}Q\ge3+2\sqrt{2}\\Q\le3-2\sqrt{2}\end{matrix}\right.\)
Vậy \(Q_{min}=3+2\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}\)
