$a)\quad y =\sin x +\cos x$
$\to y = \sqrt2\sin\left(x +\dfrac{\pi}{4}\right)$
Ta có:
$\quad -1\leqslant \sin\left(x +\dfrac{\pi}{4}\right) \leqslant 1$
$\to -\sqrt2 \leqslant \sqrt2\sin\left(x +\dfrac{\pi}{4}\right) \leqslant \sqrt2$
$\to -\sqrt2\leqslant y \leqslant \sqrt2$
Vậy $\min y =-\sqrt2;\ \max y =\sqrt2$
$b)\quad y = \sqrt3\sin x - \cos x + 2$
$\to y = 2\sin\left(x - \dfrac{\pi}{6}\right) + 2$
Ta có:
$\quad -1 \leqslant \sin\left(x +\dfrac{\pi}{6}\right) \leqslant 1$
$\to - 2\leqslant 2\sin\left(x +\dfrac{\pi}{6}\right)\leqslant 2$
$\to 0 \leqslant 2\sin\left(x +\dfrac{\pi}{6}\right)+2\leqslant 4$
$\to 0 \leqslant y \leqslant 4$
Vậy $\min y = 0;\ \max y = 4$
$c)\quad y =\dfrac{\sin x +\cos x}{\sin x -\cos x+2}$
$\to y\sin x - y\cos x + 2y = \sin x +\cos x$
$\to (y-1)\sin x - (y+1)\cos x + 2y = 0$
Phương trình có nghiệm
$\to (y-1)^2 + (y+1)^2\geqslant (2y)^2$
$\to 2y^2 + 2 \geqslant 4y^2$
$\to y^2 \leqslant 1$
$\to - 1 \leqslant y \leqslant 1$
Vậy $\min y = -1;\ \max y = 1$
