Răspuns :
[tex]Determinati \ n \in N \ pentru \ care \ numarul \ \dfrac{3}{2n+1} \ este \ natural. \\ \\ \dfrac{3}{2n+1} \in N \ daca \ 2n+1 \in D_3 \\ \\ D_3=\{1, 3\} \\ \\ 2n+1=1 \\ 2n=1-1 \\ 2n=0 \\ \boxed{n=0} \in N \\ \\ 2n+1=3 \\ 2n=3-1 \\ 2n=2 \\ \boxed{n=1} \in N \\ \\ n \in \{0,1\}[/tex]
3/(2n+1)=3 ⇒ 2n+1=3⇒ 2n=2⇒ n=1
3/(2n+1)=1 ⇒ 2n+1=1⇒ 2n=0⇒ n=0
3/(2n+1)=1 ⇒ 2n+1=1⇒ 2n=0⇒ n=0