\(\left(x-2010\right)^2\ge0\Rightarrow8.\left(x-2010\right)^2\ge0\)
\(\Rightarrow36-y^2\ge0\Rightarrow y^2\le36\Rightarrow y^2=\left\{0,1,4,9,16,25,36\right\}\)
mà \(36-y^2⋮8\Rightarrow y^2=\left\{4,36\right\}\)
TH1: \(y^2=4\Rightarrow y=\pm2\Rightarrow8.\left(x-2010\right)^2=32\Rightarrow\left(x-2010\right)^2=2^2=\left(-2\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-2010=2\\x-2010=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=2012\\x=2008\end{cases}}}\)
TH2: \(y^2=36\Rightarrow y=\pm6\Rightarrow8.\left(x-2010\right)^2=0\Rightarrow x-2010=0\Rightarrow x=2010\)
Vì x,y thuộc N => các cặp số x,y thỏa mãn là:
(2012,4);(2008,4);(2010,6)
Ta có:
36-y2=8(x-2010)2
=> 36-y2 chia hết cho 8 mà 36:8 dư 4 nên y2 chia 8 dư 4
Mà: \(y^2\le36\Rightarrow y\in\left\{0;1;2;3;4;5;6\right\}\)
Sau khi thử thì ta thấy y=2 tm
=> (x-2010)2=4
=> x-2010=2
=> x=2012
Vậy: y=2;x=2012
