Đáp án:
$\begin{array}{l}
m = 1\\
\Rightarrow y = \dfrac{{2\cos x + 2}}{{\cos x + \sin x + 2}}\\
\Rightarrow y.\cos x + y.\sin x + 2y = 2\cos x + 2\\
\Rightarrow \left( {y - 2} \right).\cos x + y.\sin x = 2 - 2y\\
\Rightarrow {\left( {y - 2} \right)^2} + {y^2} \ge {\left( {2 - 2y} \right)^2}\\
\Rightarrow {y^2} - 4y + 4 + {y^2} \ge 4{y^2} - 8y + 4\\
\Rightarrow 2{y^2} - 4y \le 0\\
\Rightarrow 2y\left( {y - 2} \right) \le 0\\
\Rightarrow 0 \le y \le 2\\
\Rightarrow \left\{ \begin{array}{l}
\max y = 2\,khi:\sin x = 0 \Rightarrow x = k\pi \\
\min y = 0\,khi:\cos x = - 1 \Rightarrow x = k2\pi
\end{array} \right.
\end{array}$
