=> 1 = 1/x + 1/y + 2/xy
=> xy/xy = y/xy + x/xy + 2/xy
=> xy/xy = (y+x+2)/xy
=> xy = y+x+2
=> xy - x - y = 2
=> xy - x - y + 1 = 3
=> (x-1)(y-1) = 3
Do x,y ∈ N* nên x-1, y-1 ∈ N
=> (x-1, y-1) = (1,3); (3,1)
=> (x,y)= (2,4); (4,2) (thử lại thỏa mãn)
Vậy (x,y)= (2,4); (4,2)