ĐKXĐ: \(x>0\)
\(\Leftrightarrow\frac{3}{2}\left(2\sqrt{x}+\frac{1}{\sqrt{x}}\right)< \frac{1}{2}\left(4x+\frac{1}{x}\right)-7\)
Đặt \(2\sqrt{x}+\frac{1}{\sqrt{x}}=t\ge2\sqrt{2}\Rightarrow4x+\frac{1}{x}=t^2-4\)
\(\frac{3}{2}t< \frac{1}{2}\left(t^2-4\right)-7\)
\(\Leftrightarrow t^2-3t-18>0\Rightarrow\left[{}\begin{matrix}t< -3\left(l\right)\\t>6\end{matrix}\right.\)
\(\Rightarrow2\sqrt{x}+\frac{1}{\sqrt{x}}>6\Leftrightarrow2x+1>6\sqrt{x}\)
\(\Leftrightarrow4x^2+4x+1>36x\)
\(\Leftrightarrow4x^2-32x+1>0\Rightarrow\left[{}\begin{matrix}0< x< \frac{8-3\sqrt{7}}{2}\\x>\frac{8+3\sqrt{7}}{2}\end{matrix}\right.\)
