Ta có : 5xy - 3 = 2xy
=> 5xy - 2xy = 3
=> 3xy = 3
=> xy = 1
=> \(\hept{\begin{cases}x=1\\y=1\end{cases}}\)(vì \(x;y\inℕ\))
5xy - 3 = 2xy
5xy - 2xy = 3
(5 - 2).xy = 3
3 xy = 3
xy = 3 : 3
xy = 1 = 1.1 mà (x ; y)\(\inℕ\)
\(\Rightarrow\hept{\begin{cases}x=1\\y=1\end{cases}}\)
Vậy x = 1 ; y = 1