Đáp án: $\left[ \begin{array}{l}
m \ge \dfrac{5}{3}\\
m \le - 1
\end{array} \right.$
Giải thích các bước giải:
$\begin{array}{l}
y = \dfrac{1}{3}\left( {{m^2} - 1} \right){x^3} - \left( {m + 1} \right){x^2} + 4x + 5\\
\Rightarrow y' = \left( {{m^2} - 1} \right).{x^2} - 2\left( {m + 1} \right).x + 4\\
y' \ge 0\forall x\\
\Rightarrow \left( {{m^2} - 1} \right){x^2} - 2\left( {m + 1} \right).x + 4 \ge 0\forall x\\
+ Khi:{m^2} - 1 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 1 \Rightarrow - 4x + 4 \ge 0\forall x\left( {ktm} \right)\\
m = - 1 \Rightarrow 4 \ge 0\forall x\left( {tm} \right)
\end{array} \right.\\
\Rightarrow chọn:m = - 1\\
+ Khi:{m^2} - 1 > 0\\
\Rightarrow \left[ \begin{array}{l}
m > 1\\
m < - 1
\end{array} \right.\\
\Rightarrow \Delta ' \le 0\\
\Rightarrow {\left( {m + 1} \right)^2} - 4.\left( {{m^2} - 1} \right) \le 0\\
\Rightarrow {m^2} + 2m + 1 - 4{m^2} + 4 \le 0\\
\Rightarrow - 3{m^2} + 2m + 5 \le 0\\
\Rightarrow 3{m^2} - 2m - 5 \ge 0\\
\Rightarrow \left( {3m - 5} \right)\left( {m + 1} \right) \ge 0\\
\Rightarrow \left[ \begin{array}{l}
m \ge \dfrac{5}{3}\\
m \le - 1
\end{array} \right.\\
Vậy\,m \ge \dfrac{5}{3}\,hoặc\,m \le - 1
\end{array}$
