\(\left(x-2013\right)^{2014}=1\\ \Leftrightarrow x-2013\in\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{2012;2014\right\}\)
Vậy....
ta có : \(\left(x-2013\right)^{2014}=1\Leftrightarrow\left[{}\begin{matrix}x-2013=\sqrt[2014]{1}\\x-2013=-\sqrt[2014]{1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2013=1\\x-2013=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2014\\x=2012\end{matrix}\right.\) vậy \(x=2014;x=2012\)
\(\left(x-2013\right)^{2014}=1\)
\(\Leftrightarrow\left(x-2013\right)^{2014}=\left(\pm1\right)^{2014}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2013=1\\x-2013=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2014\\x=2012\end{matrix}\right.\)
(x-2013)2014=1
(x-2013)2014=12014
x-2013=1
Suy ra x=1+2013
x=2014
ta có\(\left(x-2013\right)^{2014}=1\)
\(\Rightarrow x-2013=1\)
\(\Rightarrow x=2014\)
TH2
\(x-2013=-1\)
\(\Rightarrow\)x=2012
(x - 2013)2014 =1
=> (x-2013)2014 = 12014
=> x-2013 = 1
=> x= 1+2013
=> x=2014