$(2x-1)^6=(2x-1)^8$
$\leftrightarrow (2x-1)^6-(2x-1)^8=0$
$\leftrightarrow (2x-1)^6[1-(2x-1)^2]=0$
\(\leftrightarrow\left[ \begin{array}{l}(2x-1)^6=0\\1-(2x-1)^2=0\end{array} \right.\) \(\leftrightarrow \left[ \begin{array}{l}2x-1=0\\2x-1=\pm 1\end{array} \right.\) \(\left[ \begin{array}{l}2x=1\\2x=0\\2x=2\end{array} \right.\) \(\leftrightarrow\left[ \begin{array}{l}x=\dfrac{1}{2}\\x=0\\x=1\end{array} \right.\)
Vậy $x=\Big\{\dfrac{1}{2};1;0\Big\}$
