` xy - 3x + y - 8 = 0 ` `(x,y ∈ Z)`
` => x(y - 3) + (y - 3) - 5 = 0 `
` => (y - 3)(x + 1) = 5 `
` => (y - 3),(x + 1) ∈ Ư(5) = {±1 ; ±5} `
Và ` (y - 3)(x + 1) = -5.(-1) = -1.(-5) = 1.5 = 5.1 `
Ta có bảng giá trị:
\begin{array}{|c|c|c|}\hline x+1&-5&-1&1&5\\\hline y-3&-1&-5&5&1\\\hline x&-6&-2&0&4\\\hline y&2&-2&8&4\\\hline \end{array}
Vậy ` (x ; y) = (-6 ; 2),(-2 ; -2),(0 ; 8),(4 ; 4) `
