Đáp án:
\(m \in \left( { - \infty ; - \frac{5}{3}} \right] \cup \left( {1; + \infty } \right)\)
Giải thích các bước giải:
\(\begin{array}{l}
y'\left( x \right) = \left( {{m^2} - 1} \right){x^2} - 2\left( {m - 1} \right)x + 4 \ge 0\left( {ld} \right)\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
{m^2} - 1 > 0\\
{m^2} - 2m + 1 - 4\left( {{m^2} - 1} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left( {m - 1} \right)\left( {m + 1} \right) > 0\\
{m^2} - 2m + 1 - 4{m^2} + 4 \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\\
- 3{m^2} - 2m + 5 \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\\
\left( {1 - m} \right)\left( {3m + 5} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m \in \left( { - \infty ; - 1} \right) \cup \left( {1; + \infty } \right)\\
m \in \left( { - \infty ; - \frac{5}{3}} \right] \cup \left[ {1; + \infty } \right)
\end{array} \right.\\
KL:m \in \left( { - \infty ; - \frac{5}{3}} \right] \cup \left( {1; + \infty } \right)
\end{array}\)
