\(\Delta'=4-\left(-2m+1\right)=2m+5\)
Để pt có 2 nghiệm x1;x2
\(2m+5\ge0\Leftrightarrow m\ge-\dfrac{5}{2}\)
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=-4\left(1\right)\\x_1x_2=-2m+1\left(2\right)\end{matrix}\right.\)
Lại có \(x_1-3x_2=12\left(3\right)\)
Từ (1) ; (3) ta có \(\left\{{}\begin{matrix}x_1+x_2=-4\\x_1-3x_2=12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x_2=-16\\x_1=-4-x_2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=-4\\x_1=0\end{matrix}\right.\)
Thay vào (3) ta đc
\(0=-2m+1\Leftrightarrow m=\dfrac{1}{2}\left(tm\right)\)