a) \(\left(x-6\right)^{10}=\left(x-6\right)^8\)
\(\Rightarrow\left(x-6\right)^{10}-\left(x-6\right)^8=0\)
\(\Rightarrow\left(x-6\right)^8.\left[\left(x-6\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-6\right)^8=0\\\left(x-6\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-6=0\\\left(x-6\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+6\\x-6=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x-6=1\\x-6=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=7\\x=5\end{matrix}\right.\)
Vậy \(x\in\left\{6;7;5\right\}.\)
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