A={x∈N/[tex] \frac{6}{x+2} [/tex]∈N}
[tex] \frac{6}{x+2} [/tex]∈N ⇒x+2 | 6 ⇒ x+2∈D₆ (divizorii naturali ai lui 6)
D₆={1,2,3,6} vom analiza 4 cazuri
1.x+2=1⇒x=-1 (nu analizam acest caz deoarece x∈Z)
2.x+2=2⇒x=0
3.x+2=3⇒x=1
4.x+2=6⇒x=4 deci A={0;1;4}
B={x∈N/[tex] \frac{15}{2x+1} [/tex]∈N}
[tex] \frac{15}{2x+1} [/tex]∈N⇒ 2x+1 | 15 ⇔ 2x+1∈D₁₅
D₁₅={1,3,5,15}
1.2x+1=1⇒x=0
2.2x+1=3⇒x=1
3.2x+1=5⇒x=2
4.2x+1=15⇒x=7 deci B={0;1;2;7}
