\(\left(x-3\right)^7=\left(x-3\right)^2\Leftrightarrow\left(x-3\right)^7-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[\left(x-3\right)^7-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-3\right)^2=0\\\left(x-3\right)^7-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^7=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0+3=3\\x-3=1\Rightarrow x=4\end{cases}}\)
Vậy \(x\in\left\{3;4\right\}\)