Đáp án: $1-2\sqrt{2}\le y\le 1+2\sqrt2$
Giải thích các bước giải:
Ta có:
$\sin2x+\cos2x=\sqrt{2}\sin(2x+\dfrac{\pi}4)$
Vì $-1\le \sin(2x+\dfrac{\pi}4)\le 1$
$\to -\sqrt{2}\le \sqrt2\sin(2x+\dfrac{\pi}4)\le \sqrt2$
$\to -\sqrt{2}\le \sin2x+\cos2x\le \sqrt2$
$\to -2\sqrt{2}\le (\sin2x+\cos2x)^3\le 2\sqrt2$
$\to 1-2\sqrt{2}\le 1-(\sin2x+\cos2x)^3\le 1+2\sqrt2$
$\to 1-2\sqrt{2}\le y\le 1+2\sqrt2$
$\to Min y=1-2\sqrt2$ khi $\sin(2x+\dfrac{\pi}4)=1$
Và $Max y=1+2\sqrt2$ khi $\sin(2x+\dfrac{\pi}4)=-1$
