Question

[Nguyễn Thùy Linh]

Tìm min, max

y = \(2-3\sqrt{5+2cos\left(2x+\frac{\pi}{4}\right)}\)

Answer 1

[Nguyễn Minh Quang]

ta có 

\(-1\le cos\left(2x+\frac{\pi}{4}\right)\le1\Leftrightarrow-2\le2cos\left(2x+\frac{\pi}{4}\right)\le2\)

\(\Leftrightarrow5\le5+2cos\left(2x+\frac{\pi}{4}\right)\le7\Leftrightarrow\sqrt{5}\le\sqrt{5+2cos\left(2x+\frac{\pi}{4}\right)}\le\sqrt{7}\)

Vậy \(2-3\sqrt{5}\ge y\ge2-3\sqrt{7}\)

vậy \(\hept{\begin{cases}min\left(y\right)=2-3\sqrt{7}\\max\left(y\right)=2-3\sqrt{5}\end{cases}}\)