`(x+3)^2014 = (x+3)^2012`
`(x+3)^2014 -(x+3)^2012 =0`
`(x+3)^2012 [(x+3)^2 -1]=0`
TH1 :`(x+3)^2012 =0 => x+3 =0 => x=-3`
TH2 :`(x+3)^2 -1 =0 => (x+3)^2 =1 => [(x+3=1),(x+3=-1):}`
`=> [(x=-2),(x=-4):}`
`(x+3)^2014 = (x+3)^2012`
`=> (x+3)^2014 - (x+3)^2012 = 0`
`=> (x+3)^2 * (x+3)^2012 - (x+3)^2012 = 0`
`=> (x+3)^2012 * [ (x+3)^2 - 1] =0`
`=>`\(\left[{}\begin{matrix}\left(x+3\right)^{2012}=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x+3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-3\\x+3=1\\x+3=-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-3\\x=-2\\x=-4\end{matrix}\right.\)
Vậy, `x = {-3; -2; -4}.`