\(\left(2m-1\right)x^2-3mx+m-1=0\)
\(\Delta=b^2-4ac\)
\(\Delta=m^2+12m-4\)
Theo định lý Viet
\(\Rightarrow\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}\\P=x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{3m}{2m-1}\\P=x_1x_2=\dfrac{m-1}{2m-1}\end{matrix}\right.\)
Để pt có 2 nghiệm dương phân biệt
\(\Rightarrow\left\{{}\begin{matrix}\Delta>0\\S>0\\P>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m^2+12m-4>0\\\dfrac{3m}{2m-1}>0\\\dfrac{m-1}{2m-1}>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\in\left(-\infty;-6-2\sqrt{10}\right)\cup\left(-6+2\sqrt{10};+\infty\right)\\m\in\left(-\infty;0\right)\cup\left(\dfrac{1}{2};+\infty\right)\\m\in\left(-\infty;\dfrac{1}{2}\right)\cup\left(1;+\infty\right)\end{matrix}\right.\)
\(\Rightarrow m\in\left(-\infty;-6-2\sqrt{10}\right)\cup\left(1;+\infty\right)\)