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Vì $0\le\sin^2x,\cos^2x\le1$, ta có:
$11\cos^{2008}x\le11\cos^2x\le2012\cos^2x$
$2012\sin^{2010}x\le2012\sin^2x$
Suy ra: $11\cos^{2008}x+2012\sin^{2012}x\le 2012(\sin^2x+\cos^2x)=2012$
Dấu bằng xảy ra khi:
$\left\{ \begin{array}{l} \cos x=0\\\sin^2x=1 \end{array} \right.\Leftrightarrow x=\frac{\pi}{2}+k\pi,k\in\mathbb{Z}$
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