\(\frac{x^2}{9}+\frac{1}{x^2}=\frac{5}{3}\left(\frac{x}{3}-\frac{1}{x}\right)\)
đặt \(\left(\frac{x}{3}-\frac{1}{x}\right)=t\Rightarrow t^2=\left(\frac{x^2}{9}-\frac{2}{3}+\frac{1}{x^2}\right)\Rightarrow\frac{x^2}{9}+\frac{1}{x^2}=t^2+\frac{2}{3}\)
\(\Leftrightarrow t^2-\frac{5}{3}t+\frac{2}{3}=0\Leftrightarrow t^2-2.\frac{5}{6}t+\left(\frac{5}{6}\right)^2=\frac{25}{36}-\frac{24}{36}=\frac{1}{36}=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\left[\begin{matrix}t-\frac{5}{6}=\frac{1}{6}\\t-\frac{5}{6}=-\frac{1}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}t=1\\t=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\frac{x}{3}-\frac{1}{x}=1\\\frac{x}{3}-\frac{1}{x}=\frac{2}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x\ne0\\x^2-x-3=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ne0\\\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\\\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\end{matrix}\right.\)
Kết luận: \(\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\\x=-1\\x=3\end{matrix}\right.\)
