[tex] \frac{15}{x+7} [/tex] ∈ IN ⇔ x + 7 ∈ D₁₅
x + 7∈ {1, 3, 5, 15}
x ∈ {-6, -4, -2, 8}
x ∈ IN ⇒ x ∈ {8}
A = {8}
[tex] \frac{12}{2x + 3} [/tex] ∈ IN ⇔ 2x + 3 ∈ D₁₂
2x + 3 ∈ {1, 2, 3, 4, 6, 12}
2x ∈ {-2, -1, 0, 1, 3, 9}
x ∈ {-1, -1/2, 0, 1/2, 3/2, 9/2}
x ∈ IN ⇒ x ∈ {0}
B = {0}
