\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2-2m\left(x+\frac{1}{x}\right)+2m-1=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow\left[{}\begin{matrix}t\ge2\\t\le-2\end{matrix}\right.\)
\(\Rightarrow t^2-2mt+2m-1=0\)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-2m\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right)\left(t+1-2m\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=1\left(ktm\right)\\t=2m-1\end{matrix}\right.\)
Pt đã cho có nghiệm khi và chỉ khi \(\left[{}\begin{matrix}2m-1\ge2\\2m-1\le-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}m\ge\frac{3}{2}\\m\le-\frac{1}{2}\end{matrix}\right.\)