Question

Answer 1

\(\Leftrightarrow2cos^22x-2cos2x.cos\left(\dfrac{2018\pi^2}{x}\right)=2cos^22x-2\)

\(\Leftrightarrow cos2x.cos\left(\dfrac{2018\pi^2}{x}\right)=1\)

Do \(\left\{{}\begin{matrix}cos2x\le1\\cos\left(\dfrac{2018\pi^2}{x}\right)\le1\end{matrix}\right.\) \(\Rightarrow cos2x.cos\left(\dfrac{2018\pi^2}{x}\right)\le1\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}cos2x=1\\cos\left(\dfrac{2018\pi^2}{x}\right)=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=k2\pi\\\dfrac{2018\pi^2}{x}=n2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=k\pi\\x=\dfrac{1009\pi}{n}\end{matrix}\right.\) \(\Rightarrow k=\dfrac{1009}{n}\)

\(\Rightarrow n=Ư\left(1009\right)=\left\{1;1009\right\}\)

\(\Rightarrow x=\left\{\pi;1009\pi\right\}\)

Answer 2

Th2 em tịt rồi, ko bik làm sao nữa