\(25-y^2=8\left(x-2015\right)^2\)
Ta có: \(25-y^2\le25\Rightarrow8\left(x-2015\right)^2\le25\)
Mà: \(8\left(x-2015\right)^2\ge0;8\left(x-2015\right)^2⋮8\)
\(\Rightarrow\left\{{}\begin{matrix}8\left(x-2015\right)^2\in N\\8\left(x-2015\right)^2⋮8\\0\le8\left(x-2015\right)^2\le25\end{matrix}\right.\)
\(\Rightarrow8\left(x-2015\right)^2\in\left\{0;8;16;24\right\}\Rightarrow\left(x-2015\right)^2\in\left\{0;1;2;3\right\}\)
Giải tiếp nhé
\(25-y^2=8\left(x-2015\right)^2\)
\(pt\Leftrightarrow8\left(x-2015\right)^2+y^2=25\left(1\right)\)
Vì \(y^2\ge0\Rightarrow8\left(x-2015\right)^2\le25\)
\(\Rightarrow\left(x-2015\right)^2\le\dfrac{25}{8}\). Nên ta có:
*)Với \(\left(x-2015\right)^2=1\) thay vào \((1)\) ta có \(y^2=17\) (loại)
*)Với \(\left(x-2015\right)^2=0\) thay vào \((1)\) ta có \(y^2=25\Rightarrow y=\pm5\)
Vậy \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2015\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=2015\\y=-5\end{matrix}\right.\end{matrix}\right.\) thỏa mãn