\(\Delta=\left(2m-1\right)^2-4\left(m^2-1\right)=5-4m\ge0\Rightarrow m\le\dfrac{5}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1x_2=m^2-1\end{matrix}\right.\)
\(\left(x_1-x_2\right)^2=x_1-3x_2\)
\(\Leftrightarrow x_1-3x_2=\left(x_1+x_2\right)^2-4x_1x_2\)
\(\Leftrightarrow x_1-3x_2=\left(2m-1\right)^2-4\left(m^2-1\right)\)
\(\Leftrightarrow x_1-3x_2=5-4m\)
Kết hợp \(x_1+x_2=2m-1\) ta được: \(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1-3x_2=5-4m\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2m-1\\4x_2=6m-6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_2=\dfrac{3m-3}{2}\\x_1=2m-1-x_2=\dfrac{m+1}{2}\end{matrix}\right.\)
Thế vào \(x_1x_2=m^2-1\)
\(\Rightarrow\left(\dfrac{3m-3}{2}\right)\left(\dfrac{m+1}{2}\right)=m^2-1\)
\(\Leftrightarrow3\left(m^2-1\right)=4\left(m^2-1\right)\)
\(\Leftrightarrow m^2-1=0\Rightarrow m=\pm1\)
