Đáp án:
$\begin{array}{l}
y = \dfrac{{\sin x + \cos x + 3}}{{2\sin x + \cos x + 3}}\\
\Rightarrow 2y.\sin x + y.\cos x + 3y = \sin x + \cos x + 3\\
\Rightarrow \left( {2y - 1} \right).\sin x + \left( {y - 1} \right).\cos x = 3 - 3y\left( * \right)
\end{array}$
Điều kiện để pt (*) có nghiệm là:
$\begin{array}{l}
{\left( {2y - 1} \right)^2} + {\left( {y - 1} \right)^2} \ge {\left( {3 - 3y} \right)^2}\\
\Rightarrow {\left( {2y - 1} \right)^2} \ge 8{\left( {y - 1} \right)^2}\\
\Rightarrow 4{y^2} - 4y + 1 \ge 8{y^2} - 16y + 8\\
\Rightarrow 4{y^2} - 12y + 7 \le 0\\
\Rightarrow 4.\left( {{y^2} - 3y + \dfrac{9}{4}} \right) - 2 \le 0\\
\Rightarrow 4{\left( {y - \dfrac{3}{2}} \right)^2} \le 2\\
\Rightarrow {\left( {y - \dfrac{3}{2}} \right)^2} \le \dfrac{1}{2}\\
\Rightarrow \dfrac{{ - \sqrt 2 }}{2} \le y - \dfrac{3}{2} \le \dfrac{{\sqrt 2 }}{2}\\
\Rightarrow \dfrac{{3 - \sqrt 2 }}{2} \le y \le \dfrac{{3 + \sqrt 2 }}{2}\\
\Rightarrow \left\{ \begin{array}{l}
GTNN:y = \dfrac{{3 - \sqrt 2 }}{2}\\
GTLN:y = \dfrac{{3 + \sqrt 2 }}{2}
\end{array} \right.
\end{array}$
