Đáp án+Giải thích các bước giải:

$1)mx^2+(m-1)x+3(m-1)=0$

Phương trình có hai nghiệm phân biệt âm

$\Rightarrow \left\{\begin{array}{l} a \ne 0 \\  \Delta >0 \\ S<0 \\ P>0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ne 0 \\ (m-1)^2-12m(m-1)>0 \\ \dfrac{1-m}{m}<0 \\ \dfrac{3(m-1)}{m}>0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ne 0 \\ -11 m^2 + 10 m + 1 >0 \\ \dfrac{m-1}{m}>0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ne 0 \\  (1 - m) (11 m + 1) >0 \\ \left[\begin{array}{l} m>1 \\ m<0\end{array} \right. \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ne 0 \\  -\dfrac{1}{11}<m<1 \\ \left[\begin{array}{l} m>1 \\ m<0\end{array} \right. \end{array} \right.\\ \Leftrightarrow -\dfrac{1}{11}<m<0\\ 2)$

Phương trình có hai nghiệm khác $0$

$\Leftrightarrow \left\{\begin{array}{l} a \ne 0 \\ \Delta \ge 0 \\ m.0^2+(m-1).0+3(m-1) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ne 0 \\ (1 - m) (11 m + 1)  \ge 0 \\ 3(m-1) \ne 0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ne 0 \\  -\dfrac{1}{11}\le m\le 1 \\ m \ne 1\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} m \ne 0 \\  -\dfrac{1}{11}\le m < 1 \end{array} \right.\\ Vi-et: x_1+x_2=\dfrac{1-m}{m}\\ x_1x_2=\dfrac{3(m-1)}{m}\\ a) x_1+x_2=4x_1x_2+m\\ \Leftrightarrow \dfrac{1-m}{m}=\dfrac{12(m-1)}{m}+m\\ \Leftrightarrow 1-m=12(m-1)+m^2\\ \Leftrightarrow m^2 + 13 m - 13=0\\ \Leftrightarrow \left[\begin{array}{l} m=\dfrac{-13+\sqrt{221}}{2} (TM) \\m=\dfrac{-13-\sqrt{221}}{2} (L)\end{array} \right.\\ b)x_1<x_2<3\\ \Leftrightarrow \left\{\begin{array}{l} a\ne 0 \\ \Delta > 0 \\ x_1-3+x_2-3 <0 \\ (x_1-3)(x_2-3) >0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} -\dfrac{1}{11} <x<0 \\ x_1+x_2-6<0 \\ x_1x_2-3(x_1+x_2)+9 >0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} -\dfrac{1}{11}<m<0 \\ \dfrac{1-m}{m} -6<0 \\  \dfrac{3(m-1)}{m}-3.\dfrac{1-m}{m}+9>0\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} -\dfrac{1}{11}<m<0  \\\dfrac{1-7m}{m} <0 \\ \dfrac{3(m-1)}{m}+\dfrac{3(m-1)}{m}+9>0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l}-\dfrac{1}{11}<m<0 \\ \left[\begin{array}{l} m> \dfrac{1}{7} \\ m<0  \end{array} \right. \\\dfrac{6(m-1)}{m}+9>0  \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} -\dfrac{1}{11}<m<0 \\ \left[\begin{array}{l} m> \dfrac{1}{7} \\ m<0 \end{array} \right.\\ \dfrac{15m-6}{m}>0 \end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} -\dfrac{1}{11}<m<0 \\ \left[\begin{array}{l}m> \dfrac{1}{7} \\ m<0 \end{array} \right.\\ \left[\begin{array}{l} m>\dfrac{2}{5} \\ m<0 \end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l}-\dfrac{1}{11}<m<0 \\ \left[\begin{array}{l} m> \dfrac{2}{5} \\ m<0 \end{array} \right.\end{array} \right.\\ \Leftrightarrow -\dfrac{1}{11} < m<0(TM).$