[tóan 11]

H

hohungho

Last edited by a moderator:
D

demon311

Ta có:

$\cos^{13} x \le \cos^2 x \\
\sin^{12} x \le \sin^2 x \\
\rightarrow \sin^{12} x+\cos^{13} x \le 1 $
Dấu bằng xảy ra là khi pt có nghiệm
Do đó:
$\sin^{12} x=\sin^2 x \\
\cos^{13} x=\cos^2 x \\
\left[ \begin{array}{ll}
\begin{cases}
\sin x=\pm 1 \\
\cos x= 0
\end{cases} \\
\begin{cases}
\sin x=0 \\
\cos x=1
\end{cases}
\end{array} \right. \\
x=k\dfrac{ \pi}{2} \;\; (k \ne 3n \; ; \; n \in \mathbb{Z})$
 
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