24.|x-2| ≤ 1 ⇔ - 1≤ (x-2) ≤ 1 (+2) ⇒ 1 ≤ x ≤ 3 ⇒ A = {1,2,3}
(2x+1)/3x+2) ∈ Z ⇔ (3x+2)∈ D(2x+1)
daca : d | (2x+1) ⇒ d | 3(2x+1) = 6x +3 (1)
si d | (3x+2) ⇒ d | 2(3x+2) = 6x +4 (2)
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⇒ d | [(2) - (1) = 1] ⇒ d∈∅ B = ∅
A ∪ B = A = {1,2,3}
A∩B = ∅
A\B = A
B\A = ∅
A x B = B x A = ∅