Giải thích các bước giải:
\(\begin{array}{l}
\Delta ' = {m^2} - 2m + 1 - {m^2} + 3m = m + 1;\,a = 1 \ne 0\\
1)\,\Delta ' > 0 \Leftrightarrow m > - 1\\
2)\,\Delta ' \ge 0 \Leftrightarrow m \ge - 1\\
3)\,\Delta ' = 0 \Leftrightarrow m = - 1\\
{x_1} = {x_2} = m - 1 = - 2\\
4)\,Thay\,x = - 1\,vao\,pt\,ta\,duoc:\,\\
{\left( { - 1} \right)^2} - 2\left( {m - 1} \right).\left( { - 1} \right) + {m^2} - 3m = 0\\
\Leftrightarrow {m^2} - m - 1 = 0 \Leftrightarrow m = \dfrac{{1 \pm \sqrt 5 }}{2} \ge - 1\\
\Rightarrow Theo\,vi - et\,ta\,co:\\
{x_1} + {x_2} = 2\left( {m - 1} \right)\\
{x_2} = 2m - 1\\
Voi\,m = \dfrac{{1 + \sqrt 5 }}{2} \Rightarrow {x_2} = \sqrt 5 \\
Voi\,m = \dfrac{{1 - \sqrt 5 }}{2} \Rightarrow {x_2} = - \sqrt 5 \\
5)\,Theo\,Vi - et:\\
\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m - 2\\
{x_1}.{x_2} = {m^2} - 3m
\end{array} \right.\\
\Rightarrow 3\left( {{x_1} + {x_2}} \right) = 4{x_1}{x_2}\\
\Leftrightarrow 3\left( {2m - 2} \right) = 4\left( {{m^2} - 3m} \right)\\
\Leftrightarrow 4{m^2} - 18m + 6 = 0 \Leftrightarrow m = \dfrac{{9 \pm \sqrt {57} }}{4}\\
6)\,{x_1} = 3{x_2} \Rightarrow 4{x_2} = 2m - 2 \Leftrightarrow {x_2} = \dfrac{{m - 1}}{2}\\
\Rightarrow {x_1} = \dfrac{{3\left( {m - 1} \right)}}{2}\\
\Rightarrow {x_1}{x_2} = \dfrac{{3{{\left( {m - 1} \right)}^2}}}{4} = {m^2} - 3m\\
\Leftrightarrow 3{m^2} - 6m + 3 = 4{m^2} - 12m\\
\Leftrightarrow {m^2} - 6m - 3 = 0 \Leftrightarrow m = 3 \pm 2\sqrt 3
\end{array}\)
