Đáp án: $A$
Giải thích các bước giải:
Ta có:
$y=\dfrac{\cos x-2\sin x}{2-\sin x}$
$\to (2-\sin x)y=\cos x-2\sin x$
$\to 2y-y\sin x=\cos x-2\sin x$
$\to 2y=(y-2)\sin x+\cos x$
$\to (2y)^2=((y-2)\sin x+\cos x)^2$
$\to (2y)^2\le ((y-2)^2+1^2)(\sin^2x+\cos^2x)$
$\to 4y^2\le (y-2)^2+1$
$\to 3y^2+4y-5\le \:0$
$\to 3\left(y+\dfrac{2}{3}\right)^2-\dfrac{19}{3}\le \:0$
$\to \dfrac{-\sqrt{19}-2}{3}\le \:y\le \dfrac{\sqrt{19}-2}{3}$
$\to GTNN_y=\dfrac{-\sqrt{19}-2}{3}, GTLN_y= \dfrac{\sqrt{19}-2}{3}$
