10/(x-3) ∈ Z ⇔ (x-3)∈ D10 ⇒ ( x -3) ∈ {-10, -5, -2, -1, 1, 2, 5 ,10}
x ∈{-7, - 2, 1, 2, 4, 5, 8, 13} B = {1, 2, 4, 5, 8, 13}
3/(x-1) ∈ Z ⇔ (x - 1) ∈ D3 x ≠1 (x-1) ∈ {-3, -1, 3} x ∈ {-2, 0, 4}
C = {-2}
-5/(4 -x) = 5/(x-4) ∈ N ⇔ (x-4) ∈D5 (x - 4) ∈ { 1, 5} x∈ { 5, 9}
D = {5,9}