Mình sửa lại, mình đọc nhầm đề bài:
\(\left(x-5\right)^{30}-\left(x-5\right)^{28}=0\)
\(\Rightarrow\left(x-5\right)^{28}\cdot\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^{28}=0\\\left[\left(x-5\right)^2-1\right]=0\end{matrix}\right.\)
Xét \(\left(x-5\right)^{28}=0\Rightarrow x=5\)
Xét \(\left(x-5\right)^2-1=0\)
\(\Rightarrow\left[{}\begin{matrix}x-5=1\Rightarrow x=6\\x-5=-1\Rightarrow x=4\end{matrix}\right.\)
Vậy \(x\in\left\{4;5;6\right\}\)
