Đáp án:
$\begin{array}{l}
\dfrac{{2\sin x + \cos x + 1}}{{\sin x - 2\cos x + 3}} = m\\
\Leftrightarrow 2\sin x + \cos x + 1 = m\sin x - 2m\cos x + 3m\\
\Leftrightarrow \left( {2 - m} \right).\sin x + \left( {1 + 2m} \right).\cos x = 3m - 1\\
Dk:{\left( {2 - m} \right)^2} + {\left( {1 + 2m} \right)^2} \ge {\left( {3m - 1} \right)^2}\\
\Leftrightarrow 4 - 4m + {m^2} + 1 + 4m + 4{m^2} \ge 9{m^2} - 6m + 1\\
\Leftrightarrow 4{m^2} - 6m - 4 \le 0\\
\Leftrightarrow 2{m^2} - 3m - 2 \le 0\\
\Leftrightarrow 2{m^2} - 4m + m - 2 \le 0\\
\Leftrightarrow \left( {m - 2} \right)\left( {2m + 1} \right) \le 0\\
\Leftrightarrow - \dfrac{1}{2} \le m \le 2\\
Vậy\, - \dfrac{1}{2} \le m \le 2\\
b)\sin x + m\cos x = m\\
\Leftrightarrow {1^2} + {m^2} \ge {m^2}\\
\Leftrightarrow 1 \ge 0\left( {tm} \right)\\
Vậy\,m \in R
\end{array}$
