\(\Delta'=\left(m-1\right)^2-\left(3-4m\right)=m^2+2m-2\ge0\) \(\Rightarrow\left[{}\begin{matrix}m\ge\sqrt{3}-1\\m\le-\sqrt{3}-1\end{matrix}\right.\)
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m-1\right)\left(1\right)\\x_1x_2=3-4m\left(2\right)\end{matrix}\right.\)
\(x_1=3x_2\) thay vào (1):
\(4x_2=2\left(m-1\right)\Rightarrow x_2=\frac{m-1}{2}\Rightarrow x_1=\frac{3\left(m-1\right)}{2}\) thay vào (2):
\(\left(\frac{m-1}{2}\right)\left(\frac{3\left(m-1\right)}{2}\right)=3-4m\)
\(\Leftrightarrow3\left(m-1\right)^2-4\left(3-4m\right)=0\)
\(\Leftrightarrow3m^2+10m-9=0\Rightarrow\left[{}\begin{matrix}m=\frac{-5+2\sqrt{13}}{3}\\m=\frac{-5-2\sqrt{13}}{3}\end{matrix}\right.\)
