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Ta có:
$y=\frac{\cos x+2}{\cos x+\sin x-2}$
$\Leftrightarrow y(\sin x+\cos x-2)=\cos x+2$
$\Leftrightarrow y\sin x+(y-1)\cos x=2y+2$
Suy ra:
$(2y+2)^2=[y\sin x+(y-1)\cos x]^2$
$\le [y^2+(y-1)^2][\sin^2x+\cos^2x]$
$\Rightarrow (2y+2)^2\le y^2+(y-1)^2$
$\Leftrightarrow 2y^2+10y+3\le0$
$\Leftrightarrow \frac{-5-\sqrt{19}}{2}\le y\le\frac{-5+\sqrt{19}}{2}$
Vậy: Min$y=\frac{-5-\sqrt{19}}{2}$
Max$y=\frac{-5+\sqrt{19}}{2}$
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