$y=\dfrac{2\cos x+2}{\cos x+\sin x+2}$ TXĐ: $D=R$
$\Leftrightarrow y(\sin x +\cos x +2) =2\cos x +2$
$\Leftrightarrow y\sin x +(y-2)\cos x = 2-2y$
Pt có nghiệm $\Leftrightarrow y^2+(y-2)^2 \ge (2-2y)^2$
$\Leftrightarrow y^2-2y \le 0$
$\Leftrightarrow 0\le y\le 2$
Vậy $\max y = 2;\ \min y = 0$