Vì \(\left(x-2015\right)^{2014}\ge0;\left(x-2016\right)^{2014}\ge0\)
=> \(\left(x-2015\right)^{2014}+\left(x-2016\right)^{2014}\ge0\)
Mà x - 2015 > x - 2016 => \(\left(x-2015\right)^{2014}>\left(x-2016\right)^{2014}\)
=> (x - 2015)2014 = 1;(x - 2016)2014 = 0
=> x - 2016 = 0
=> x = 2016
Đặt \(x-2015=a;\text{ }2016-x=b\)
\(\Rightarrow a+b=1\text{ }\left(1\right)\)
Từ phương trình đã cho, ta được \(a^{2014}+b^{2014}=1\text{ }\left(2\right)\)
Nếu \(a< 0\), \(\left(1\right)\Rightarrow b=1-a>1\), \(\Rightarrow a^{2014}+b^{2014}>1\)(không thỏa (2))
Tương tự với b
Vậy \(a,b\ge0\)
\(\left(2\right)\Rightarrow a^{2014};\text{ }b^{2014}\le1\Rightarrow-1\le a,b\le1\)
\(\Rightarrow0\le a,b\le1\)
\(\left(1\right)+\left(2\right)\Rightarrow a+b=a^{2014}+b^{2014}\)
\(\Leftrightarrow a\left(1-a^{2013}\right)+b\left(1-b^{2013}\right)=0\text{ }\left(3\right)\)
Ta lại có: \(0\le a,b\le1\Rightarrow\hept{\begin{cases}1-a^{2013}\ge0\\1-b^{2013}\ge0\end{cases}}\)
\(\Rightarrow a\left(1-a^{2013}\right)+b\left(1-b^{2013}\right)\ge0\forall a,b\in\left[0;1\right]\)
Dấu bằng chỉ xảy ra khi \(a,b\in\left\{0;1\right\}\)
Do \(a+b=1\) nên \(\left(a;b\right)\in\left\{\left(0;1\right);\text{ }\left(1;0\right)\right\}\)
+TH1: \(\hept{\begin{cases}a=1\\b=0\end{cases}}\Rightarrow\hept{\begin{cases}x-2015=1\\2016-x=0\end{cases}}\Leftrightarrow x=2016\)
+TH2 \(\hept{\begin{cases}a=0\\b=1\end{cases}\Leftrightarrow\hept{\begin{cases}x-2015=0\\2016-x=1\end{cases}}\Leftrightarrow}x=2015\)
Vậy \(x\in\left\{2015;\text{ }2016\right\}\)
Vì (x−2015)^2014≥0;(x−2016)^2014≥0
=> (x−2015)^2014+(x−2016)^2014≥0
Mà x - 2015 > x - 2016 => (x−2015)^2014>(x−2016)^2014
=> (x - 2015)^2014 = 1;(x - 2016)^2014 = 0
=> x - 2016 = 0
=> x = 2016