Đáp án:
Giải thích các bước giải:
\[\begin{array}{l}
y' = 3{x^2} - 2\left( {m - 1} \right)x + m\\
\Delta ' = {\left( {m - 1} \right)^2} - 3m = {m^2} - 5m + 1 > 0 \Leftrightarrow \left[ \begin{array}{l}
m > \frac{{5 + \sqrt {21} }}{2}\\
m < \frac{{5 - \sqrt {21} }}{2}
\end{array} \right.\\
Viet\,\,\,\,\left\{ \begin{array}{l}
{x_1} + {x_2} = \frac{{2\left( {m - 1} \right)}}{3}\\
{x_1}{x_2} = \frac{m}{3}
\end{array} \right.\\
Co\,\,\,{x_1} + 2{x_2} = - 1 \Rightarrow \left( {{x_1} + {x_2}} \right) + {x_2} = - 1 \Leftrightarrow \frac{{2\left( {m - 1} \right)}}{3} + {x_2} = - 1\\
\Leftrightarrow 2m - 2 + 3{x_2} = - 3 \Leftrightarrow {x_2} = \frac{{ - 1 - 2m}}{3} \Rightarrow {x_1} = \frac{{4m - 1}}{3}\\
\Rightarrow {x_1}{x_2} = \frac{m}{3} \Leftrightarrow \frac{{4m - 1}}{3} \cdot \frac{{ - 1 - 2m}}{3} = \frac{m}{3}\\
\Leftrightarrow - 8{m^2} - 2m + 1 = 3m \Leftrightarrow - 8{m^2} - 5m + 1 = 0 \Leftrightarrow \left[ \begin{array}{l}
m = \frac{{5 + \sqrt {57} }}{{ - 16}}\\
m = \frac{{5 - \sqrt {57} }}{{ - 16}}
\end{array} \right.\,\,\,\left( {TM} \right)\\
Vay\,\,m = \frac{{ - 5 \pm \sqrt {57} }}{{16}}
\end{array}\]
