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

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}$.