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giải phương trình : sin^8x+cos^8x = 1/8
thanks cac ban truoc nha
$\begin{array}{l}
{\sin ^8}x + {\cos ^8}x = \frac{1}{8}\\
\Leftrightarrow {\left( {{{\sin }^2}x} \right)^3} + {\left( {{{\cos }^2}x} \right)^3} = \frac{1}{8}\\
\Leftrightarrow \left( {{{\sin }^2}x + {{\cos }^2}x} \right)\left( {{{\sin }^4}x + {{\cos }^4}x - {{\sin }^2}x{{\cos }^2}x} \right) = \frac{1}{8}\\
\Leftrightarrow {\sin ^4}x + {\cos ^4}x - {\sin ^2}x{\cos ^2}x = \frac{1}{8}\\
\Leftrightarrow {\left( {{{\sin }^2}x + {{\cos }^2}x} \right)^2} - 2{\sin ^2}x{\cos ^2}x - {\sin ^2}x{\cos ^2}x = \frac{1}{8}\\
\Leftrightarrow 1 - 3{\sin ^2}x{\cos ^2}x = \frac{1}{8}
\end{array}$
tới đây dễ rồi nhé bạn.