Question

[Khuất Đăng Mạnh]

Tìm x,y \(\in\) N biết :25-\(y^2=8\cdot\left(x-2009\right)^2\)

Answer 1

[Nguyễn Huy Tú]

\(25-y^2=8\left(x-2009\right)^2\)

Ta có: \(25-y^2\le25\)

\(\Rightarrow8\left(x-2009\right)^2\le25\)

\(\Rightarrow\left(x-2009\right)^2< 4\)

Do \(x\in N\Rightarrow\left[{}\begin{matrix}\left(x-2009\right)^2=1\\\left(x-2009\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2010\\x=2009\end{matrix}\right.\left(x\in N\right)\)

+) Xét x = 2010

\(\Rightarrow25-y^2=8\Rightarrow y^2=17\) ( loại )

+) Xét x = 2009

\(\Rightarrow25-y^2=0\Rightarrow y=5\left(y\in N\right)\)

Vậy x = 2009, y = 5