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Ta có:
$y = \frac{\sin x + \cos x -1 }{\sin x -\cos x +3 }$
$\Leftrightarrow y(\sin x-\cos x+3)=\sin x+\cos x-1$
$\Leftrightarrow 3y+1=(1-y)\sin x+(y+1)\cos x$
$\Rightarrow (3y+1)^2=[(1-y)\sin x+(y+1)\cos x]^2\le[(1-y)^2+(y+1)^2][\sin^2x+\cos^2x]$
$\Rightarrow
(3y+1)^2
\le(1-y)^2+(y+1)^2$
$\Leftrightarrow 7y^2+6y-1\le 0\Leftrightarrow -1\le y\le\frac{7}{6}$
Vậy: Min$y=-1 $, Max$y=\frac{7}{6}$.
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