Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{1}{x}+1}-\frac{2}{x+\frac{1}{x}-1}=-\frac{7}{3}\)
Đặt \(x+\frac{1}{x}-1=t\)
\(\Rightarrow\frac{3}{t+2}-\frac{2}{t}=-\frac{7}{3}\)
\(\Leftrightarrow9t-6\left(t+2\right)=-7t\left(t+2\right)\)
\(\Leftrightarrow7t^2+17t-12=0\Rightarrow\left[{}\begin{matrix}t=-3\\t=\frac{4}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{x}-1=-3\\x+\frac{1}{x}-1=\frac{4}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\7x^2-11x+7=0\end{matrix}\right.\) (casio)
