Đáp án:
Không có m.
Cách làm:
$\begin{array}{l}
y' = 6{x^2} + 6\left( {m - 1} \right)x + 6\left( {m - 2} \right) = 6\left[ {{x^2} + \left( {m - 1} \right)x + m - 2} \right]\\
= 6\left( {x + 1} \right)\left( {x + m - 2} \right)\\
y' = 0 \Leftrightarrow \left[ \begin{array}{l}
x = - 1\\
x = 2 - m
\end{array} \right.\\
Hs\,co\,CD,CT\,thuoc\,khoang\,\left( {2;3} \right) \Leftrightarrow \left\{ \begin{array}{l}
- 1 \ne 2 - m\\
- 1 \in \left( {2;3} \right)\,\,\left( {vo\,ly} \right)\\
2 - m \in \left( {2;3} \right)
\end{array} \right.\\
Vay\,khong\,co\,m\,thoa\,man.
\end{array}$
