Question

multimea A={n apartine N / 3n+17 supra n+2 apartine N}

Answer 1

    
(3n + 17) / (n + 2) = (3n + 6 + 11) / (n+ 2) = (3n + 6) / (n + 2) + 11 / (n + 2) =

3(n + 2) / (n + 2) + 11 / (n + 2) = 3 + 11 / (n + 2)

11 se divide doar cu 1 si cu 11.

n + 2 = 1   =>   n = 1 - 2 = - 1 (Solutie eliminata deoarece n nu poate fi negativ)

n + 2 = 11  =>  n = 11 - 2 = 9  (Solutie acceptata deoarece n = 9 ∈ N)

=> n = 9   solutie unica.

=> A = {9}

Proba:
(3n + 17) / (n + 2) = (3 × 9 + 17) / (9 + 2) = (27+17) / (11) = 44 / 11 = 4 ∈ N

Answer 2

A={n ∈ N | $\frac{3n+17}{n+2}$∈ N}

$\frac{3n+17}{n+2} =\frac{3n+6}{n+2}+\frac{11}{n+2}=\frac{3(n+2)}{n+2}+\frac{11}{n+2} =3+\frac{11}{n+2}$

3∈N, ramane ca $\frac{11}{n+2}$ ∈N=>
n+2=11=> n=9