Đáp án: $m \ge \frac{3}{2}$
Giải thích các bước giải:
$\begin{array}{l}
y' = - 3{x^2} + 2\left( {m + 1} \right)x + 2 \ge 0\,\,\,\forall x \in \left[ {0;2} \right]\\
\Rightarrow 3{x^2} - 2 \le 2\left( {m + 1} \right)x\,\,\,\,\forall x \in \left[ {0;2} \right]\\
+ thay\,x = 0\,vao\,bpt\,ta\,duoc: - 2 \le 0\,\left( {tm} \right)\\
+ x \ne 0\,bpt \Rightarrow 3x - \frac{2}{x} \le 2\,\left( {m + 1} \right)\,\,\,\forall x \in (0;2]\\
goi\,g\left( x \right) = 3x - \frac{2}{x}\,tren\,\,(0;2]\\
g'\left( x \right) = 3 + \frac{2}{{{x^2}}} > 0\,\,\forall x \in (0;2]\\
\Rightarrow g\left( x \right) \le 2\,\left( {m + 1} \right)\,\,\,\forall x \in (0;2]\\
\Leftrightarrow m{\rm{axg}}\left( x \right) \le 2\left( {m + 1} \right)\\
\Rightarrow g\left( 2 \right) \le 2\left( {m + 1} \right)\\
\Rightarrow 3.2 - \frac{2}{2} \le 2\left( {m + 1} \right) \Rightarrow 5 \le 2\left( {m + 1} \right) \Rightarrow m \ge \frac{3}{2}\\
VAy\,m \ge \frac{3}{2}\,thi\,tmyc
\end{array}$
