B={x∈Z| -5<x-1 ≤3} ⇔B={x∈Z| -4<x ≤4} ⇔B={-3, -2, -1, 0, 1, 2, 3, 4}
15/(3x+1) ∈Z, x∈Z ⇔(3x+1) divizor al lui 15, x∈Z ⇔ (3x+1) ∈ {-15, -5,-3,-1,1,3,5,15}∩Z⇔(3x) ∈ {-16, -6,-4,-2,0,2,4,14}∩Z ⇔ x ∈ {-16:3, -6:3,-4:3,-2:3,0:3,2:3,4:3,14:3}∩Z ⇔x ∈ {-2, 0}⇒ A= {-2, 0}
A∩B ={-2, 0}
raspuns: A∩B ={-2, 0}
Răspuns:
B={x∈Z| -5<x-1 ≤3} ⇔B={x∈Z| -4<x ≤4} ⇔B={-3, -2, -1, 0, 1, 2, 3, 4}
15/(3x+1) ∈Z, x∈Z ⇔(3x+1) divizor al lui 15, x∈Z ⇔ (3x+1) ∈ {-15, -5,-3,-1,1,3,5,15}∩Z⇔(3x) ∈ {-16, -6,-4,-2,0,2,4,14}∩Z ⇔ x ∈ {-16:3, -6:3,-4:3,-2:3,0:3,2:3,4:3,14:3}∩Z ⇔x ∈ {-2, 0}⇒ A= {-2, 0}
A∩B ={-2, 0}
raspuns: A∩B ={-2, 0}
Explicație pas cu pas: