Đáp án:
\[m < - 2\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a\,{x^2} + bx + c > 0,\,\,\,\forall x \in R \Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta < 0
\end{array} \right.\\
f\left( x \right) = \frac{1}{3}\left( {{m^2} - m - 6} \right){x^3} + \left( {m + 2} \right){x^2} + x - m\\
\Rightarrow f'\left( x \right) = \left( {{m^2} - m - 6} \right){x^2} + 2\left( {m + 2} \right)x + 1\\
f'\left( x \right) > 0,\,\,\,\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} - m - 6 > 0\\
\Delta ' = {\left( {m + 2} \right)^2} - \left( {{m^2} - m - 6} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {m - 3} \right)\left( {m + 2} \right) > 0\\
\left( {{m^2} + 4m + 4} \right) - \left( {{m^2} - m - 6} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 3\\
m < - 2
\end{array} \right.\\
5m + 10 < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m > 3\\
m < - 2
\end{array} \right.\\
m < - 2
\end{array} \right. \Leftrightarrow m < - 2
\end{array}\)
Vậy \(m < - 2\)
