Răspuns :
A={x ∈ N/ 8pe x-3 ∈ N}
8pe x-3∈N⇒(x-3)∈divizorilor lui 8 ⇒ (x-3)∈{1; 2; 4; 8}⇒
x-3=1⇒ x=4
x-3=2⇒ x=5
x-3=4⇒ x=7
x-3=8⇒ x=11
x∈{4; 5; 7; 11}
8pe x-3∈N⇒(x-3)∈divizorilor lui 8 ⇒ (x-3)∈{1; 2; 4; 8}⇒
x-3=1⇒ x=4
x-3=2⇒ x=5
x-3=4⇒ x=7
x-3=8⇒ x=11
x∈{4; 5; 7; 11}
[tex] \frac{8}{x-3} [/tex] ∈ N => x-3|x-3 | * 8 && x-3|8 | *x <=> x-3|8x-24 && x-3| 8x "-"
=> x-3 ∈ [tex]D_{24} [/tex] => x-3 ∈ { 1,2,3,4,6,8,12,24} => x ∈ { 4,5,6,7,8,11,15,27}, dar x trebuie sa indeplineasca: [tex] \frac{8}{x-3} [/tex] ∈ N => x ∈ {4,5,7,11}
=> x-3 ∈ [tex]D_{24} [/tex] => x-3 ∈ { 1,2,3,4,6,8,12,24} => x ∈ { 4,5,6,7,8,11,15,27}, dar x trebuie sa indeplineasca: [tex] \frac{8}{x-3} [/tex] ∈ N => x ∈ {4,5,7,11}