Ta có: \(36-y^2\le36\)
\(\Rightarrow12\left(x-2015\right)^2\le36\)
\(\Rightarrow\left(x-2015\right)^2\le3\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2015\right)^2=0\\\left(x-2015\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}36-y^2=0\\36-y^2=1\end{matrix}\right.\Rightarrow y^2=36\Rightarrow y=6\left(y\in N\right)\)
\(\left(x-2015\right)^2=0\Rightarrow x=2015\)
Vậy x = 2015 và y = 6
\(12-\left(x-2015\right)^2=36-y^2\)
\(y^2\ge0\Rightarrow36-y^2\le36\)
\(\Rightarrow12\left(x-2015\right)^2\le36\)
\(\Rightarrow\left(x-2015\right)^2\le3\)
Mà: \(\left(x-2015\right)^2\ge0\)
\(\Rightarrow0\le\left(x-2015\right)^2\le3\)
Mà:
\(x;y\in N\) \(;\left(x-2015\right)^2\) là số chính phương nên:
\(\left(x-2015\right)^2=\left\{0;1\right\}\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x-2015=0\Rightarrow x=2015\\\left\{{}\begin{matrix}x-2015=1\Rightarrow x=2016\\x-2015=-1\Rightarrow2=2014\end{matrix}\right.\end{matrix}\right.\)
Với:
\(\left(x-2015\right)^2=1\Rightarrow36-y^2=1\) \(\Rightarrow y^2=35\Rightarrow y=\pm\sqrt{35}\) (loại vì ko tm đk)
\(\Rightarrow36-y^2=0\Rightarrow y^2=36\Rightarrow y=6\)
Vậy xảy ra khi
\(x=2015;y=6\)
