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}$.
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}$.