Đáp án:
m=2
Giải thích các bước giải:
Để phương trình có 2 nghiệm
\(\begin{array}{l}
\to \Delta \ge 0\\
\to {m^2} + 2m + 1 - 4\left( { - 3m + 2} \right) \ge 0\\
\to {m^2} + 2m + 1 + 12m - 8 \ge 0\\
\to {m^2} + 14m - 7 \ge 0\\
\to {m^2} + 14m + 49 - 56 \ge 0\\
\to {\left( {m + 7} \right)^2} \ge 56\\
\to \left[ \begin{array}{l}
m + 7 \ge 2\sqrt {14} \\
m + 7 \le - 2\sqrt {14}
\end{array} \right.\\
\to \left[ \begin{array}{l}
m \ge - 7 + 2\sqrt {14} \\
m \le - 7 - 2\sqrt {14}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{m + 1 + \sqrt {{m^2} + 14m - 7} }}{2}\\
x = \dfrac{{m + 1 - \sqrt {{m^2} + 14m - 7} }}{2}
\end{array} \right.\\
Vi - et:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 1\\
{x_1}{x_2} = - 3m + 2
\end{array} \right.\\
3{x_1} + 2{x_2} = 5\\
\to 2\left( {{x_1} + {x_2}} \right) + {x_1} = 5\\
\to 2\left( {m + 1} \right) + {x_1} = 5\\
\to \left[ \begin{array}{l}
2m + 2 + \dfrac{{m + 1 + \sqrt {{m^2} + 14m - 7} }}{2} = 5\\
2m + 2 + \dfrac{{m + 1 - \sqrt {{m^2} + 14m - 7} }}{2} = 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
4m + 4 + m + 1 + \sqrt {{m^2} + 14m - 7} = 10\\
4m + 4 + m + 1 - \sqrt {{m^2} + 14m - 7} = 10
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt {{m^2} + 14m - 7} = 5 - 5m\\
\sqrt {{m^2} + 14m - 7} = 5m - 5
\end{array} \right.\\
\to \sqrt {{m^2} + 14m - 7} = 5m - 5\\
\to {m^2} + 14m - 7 = 25{m^2} - 50m + 25\left( {DK:m \ge 1} \right)\\
\to 24{m^2} - 64m + 32 = 0\\
\to \left[ \begin{array}{l}
m = 2\left( {TM} \right)\\
m = \dfrac{2}{3}\left( l \right)
\end{array} \right.
\end{array}\)
