Giải:
Ta có:
\(\left(x-2\right)^{2016}\ge0;\forall x\)
\(\left(x-2\right)^{2018}\ge0;\forall x\)
\(\Rightarrow\left(x-2\right)^{2016}=\left(x-2\right)^{2018}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)
Vậy ...
\(\left(x-2\right)^{2016}=\left(x-2\right)^{2018}\)
\(\Leftrightarrow\left(x-2\right)^{2016}-\left(x-2\right)^{2018}=0\)
\(\Leftrightarrow\left(x-2\right)^{2016}\left[1-\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow-\left(x-2\right)^{2016}\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=3\end{matrix}\right.\)
Vậy.................
