Đáp án: $m \in \left[ { - 2;0} \right] \cup \left[ {\dfrac{1}{2};\dfrac{5}{2}} \right]$
Giải thích các bước giải:
$\begin{array}{l}
5.\cos \left( {5x + 1} \right) = 2{m^2} - m - 5\\
\Leftrightarrow \cos \left( {5x + 1} \right) = \dfrac{{2{m^2} - m - 5}}{5}\\
Do: - 1 \le \cos \left( {5x + 1} \right) \le 1\\
\Leftrightarrow - 1 \le \dfrac{{2{m^2} - m - 5}}{5} \le 1\\
\Leftrightarrow \left\{ \begin{array}{l}
\dfrac{{2{m^2} - m - 5}}{5} + 1 \ge 0\\
\dfrac{{2{m^2} - m - 5}}{5} - 1 \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
2{m^2} - m - 5 + 5 \ge 0\\
2{m^2} - m - 10 \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m\left( {2m - 1} \right) \ge 0\\
\left( {2m - 5} \right)\left( {m + 2} \right) \le 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
\left[ \begin{array}{l}
m \ge \dfrac{1}{2}\\
m \le 0
\end{array} \right.\\
- 2 \le m \le \dfrac{5}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2} \le m \le \dfrac{5}{2}\\
- 2 \le m \le 0
\end{array} \right.\\
Vậy\,m \in \left[ { - 2;0} \right] \cup \left[ {\dfrac{1}{2};\dfrac{5}{2}} \right]
\end{array}$
