Đáp án:
Pt có 2 nghiệm phân biệt thì:
$\begin{array}{l}
\left\{ \begin{array}{l}
m + 1 \ne 0\\
\Delta > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
{\left( {3m - 1} \right)^2} - 4\left( {m + 1} \right).\left( {2m - 2} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
9{m^2} - 6m + 1 - \left( {4m + 4} \right)\left( {2m - 2} \right) > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
9{m^2} - 6m + 1 - 8{m^2} + 8 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
{m^2} - 6m + 9 > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
{\left( {m - 3} \right)^2} > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \ne - 1\\
m \ne 3
\end{array} \right.\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{{1 - 3m}}{{m + 1}}\\
{x_1}{x_2} = \dfrac{{2m - 2}}{{m + 1}}
\end{array} \right.\\
Khi:{x_1} + {x_2} = 3\\
\Leftrightarrow \dfrac{{1 - 3m}}{{m + 1}} = 3\\
\Leftrightarrow 1 - 3m = 3m + 3\\
\Leftrightarrow 6m = - 2\\
\Leftrightarrow m = - \dfrac{1}{3}\\
\Leftrightarrow \left( { - \dfrac{1}{3} + 1} \right){x^2} + \left( {3.\dfrac{{ - 1}}{3} - 1} \right).x + 2.\dfrac{{ - 1}}{3} - 2 = 0\\
\Leftrightarrow \dfrac{2}{3}{x^2} - 2x - \dfrac{8}{3} = 0\\
\Leftrightarrow {x^2} - 3x - 4 = 0\\
\Leftrightarrow \left( {x - 4} \right)\left( {x + 1} \right) = 0\\
\Leftrightarrow x = 4;x = - 1\\
Vậy\,m \ne - 1;m \ne 3.Khi:{x_1} + {x_2} = 3 \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
x = - 1
\end{array} \right.
\end{array}$
