\(\Delta\) = \(\left(2m-1\right)^2-4\left(m^2-1\right)\) \(\Leftrightarrow\) \(4m^2-4m+1-4m^2+4\)
\(\Leftrightarrow\) \(5-4m\)
phương trình có 2 nghiệm \(\Leftrightarrow\) \(\Delta\) \(>0\) \(\Leftrightarrow\) \(5-4m\ge0\) \(\Leftrightarrow\) \(4m\le5\) \(m\le\dfrac{5}{4}\)
theo hệ thức vi ét ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1x_2=m^2-1\end{matrix}\right.\)
ta có : \(\left(x_1-x_2\right)^2=x_1-3x_2\) \(\Leftrightarrow\) \(\left(x_1+x_2\right)^2-4x_1x_2=x_1-3x_2\)
\(\Leftrightarrow\) \(x_1-3x_2=\left(2m-1\right)^2-4\left(m^2-1\right)\)
\(\Leftrightarrow\) \(x_1-3x_2=4m^2-4m+1-4m^2+4\)
\(\Leftrightarrow\) \(x_1-3x_2=5-4m\)
ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m-1\\x_1-3x_2=5-4m\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}4x_2=6m-6\\x_1+x_2=2m-1\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}4x_2=6m-6\\4x_1+4x_2=8m-4\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}4x_2=6m-6\\4x_1+6m-6=8m-4\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left\{{}\begin{matrix}4x_2=6m-6\\4x_1=2m+2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}x_2=\dfrac{6m-6}{4}\\x_1=\dfrac{2m+2}{4}\end{matrix}\right.\)
ta có : \(x_1x_2=m^2-1\) \(\Leftrightarrow\) \(\dfrac{\left(6m-6\right)\left(2m+2\right)}{16}=m^2-1\)
\(\Leftrightarrow\) \(\dfrac{12m^2+12m-12m-12}{16}=m^2-1\) \(\Leftrightarrow\) \(\dfrac{12m^2-12}{16}=m^2-1\)
\(\Leftrightarrow\) \(12m^2-12=16\left(m^2-1\right)\) \(\Leftrightarrow\) \(12m^2-12=16m^2-16\)
\(\Leftrightarrow\) \(4m^2-4=0\) \(\Leftrightarrow\) \(4m^2=4\) \(\Leftrightarrow\) \(m^2=1\) \(\Leftrightarrow\) \(m=\pm1\) (tmđk)
vậy \(m=\pm1\) thì \(\left(x_1-x_2\right)^2=x_1-3x_2\)
