pt <=> \(\left(3x-1\right)^{2013}-\left(3x-1\right)^{2015}=0\)
\(\Leftrightarrow\left(3x-1\right)^{2013}\left(\left(3x-1\right)^2-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)^{2013}\left(3x-2\right)3x=0\)
\(\orbr{\begin{cases}x=0\\3x-1=0,3x-2=0\end{cases}}\)
Vậy x=0, x=1/3,hoặc x=2/3