\(\left(\dfrac{2x^5-1}{3}\right)^{x^2-2x}=1\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x^2-2x=0\\\dfrac{2x^5-1}{3}\ne0\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-2x=2k\\\dfrac{2x^5-1}{3}=\pm\end{matrix}\right.\\\left\{{}\begin{matrix}x^2-2x=2k+1\\\dfrac{2x^5-1}{3}=1\end{matrix}\right.\end{matrix}\right.\)
(I) \(\left\{{}\begin{matrix}x^2-2x=0\\\dfrac{2x^5-1}{3}\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\\dfrac{2\left\{0;2\right\}^5-1}{3}\ne0\end{matrix}\right.\) =>x={0,2} là No.
(II) \(\left[{}\begin{matrix}\dfrac{2x^5-1}{3}=1\Rightarrow x^5=4\left(loai\right)\\\dfrac{2x^5-1}{3}=-1\Rightarrow x^5=-1\end{matrix}\right.\) với x=-1 có {1+2=3=> loại
Kết luận: có 2 nghiệm x={0,2}
