\(VT\ge0\Rightarrow\)\(-5\le y\le5\)

\(VT=8k^2\Rightarrow25-y^2=8k^2\Rightarrow k^2\le3\)
\(k^2=\left\{0,1\right\}\)

\(k=0\Rightarrow\hept{\begin{cases}x=2009\\y=+-5\end{cases}}\)

\(k^2=1\Rightarrow y^2=17\left(loai\right)\)

KL

\(\left(x,y\right)=\left(2009,-5\right);\left(2009,5\right)\)

Ta có: 8.(x-2013)²+y²=25

=>y²=25-8.(x-2013)²

Vì \(\left(x-2013\right)^2\ge0=>8.\left(x-2013\right)^2\ge0=>25-8.\left(x-2013\right)^2\le25-0\)

=>\(y^2\le25=>y\le5\)

=>\(y\in\left\{1,2,3,4,5\right\}=>y^2\in\left\{1,4,9,16,25\right\}\)

Vì 25:8 dư 1, 8.(x-2013)² chia 8 dư 0

=>25-8.(x-2013)² chia 8 dư 1

=>y² chia 8 dư 1

mà \(y^2\in\left\{1,4,9,16,25\right\}\)

=>y²=25=>y=5

25-8.(x-2013)²=25

=>8.(x-2013)²=0

=>(x-2013)²⁼⁰

=>x-2013=0

=>x=2013

Vậy x=2013, y=5

\(\left(x-2016\right)^2\ge0\Rightarrow5-y^2\ge0\Rightarrow y^2\le5\Rightarrow y^2\in\left\{0,1,4\right\}\)

\(\Rightarrow5-y^2\in\left\{5,4,1\right\}\)Mà x,y là STN 

\(\Rightarrow5-y^2=4\Rightarrow y^2=1\Rightarrow y=1\)

\(\Rightarrow\left(x-2016\right)^2=4\Rightarrow\orbr{\begin{cases}x-2016=2\\x-2016=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=2018\\x=2014\end{cases}}}\)

Vậy ...

vế phải luôn lẻ mọi Y

vế trái phải lẻ vậy x=0 (duy nhất chưa đủ)

1+625=5^y

625=5^?=5^y

y=?

Nếu\(x\in\)N* thì 2^x chẵn mà 624 chẵn nên 2^x + 624 hay 5^y chẵn (vô lý vì 5^y lẻ)

=> x\(\notin\)N* ; x = 0 => 5^y = 2⁰ + 624 = 1 + 624 = 625 = 5³ => y = 3

Vậy x = 0 ; y = 3

x,y=(0,4)