Question

Determinati elementele multimii A= {n ∈ N | 12/n+3(12 supra n+3) ∈ N} ajutoor

Answer 1

[tex]A=\{n \in N| \frac{12}{n+3} \in N \} \\ n+3|12 \\ n+3 \in \{1,2,3,4,6,12\} \\ n \in \{-2,-1,0,3,9\} \\ n \in N \Longrightarrow A=\{0,3,9\}[/tex]

Answer 2

rebuie ca 12/(n+3)sa fie nr natural.deci rezulta ca (n+3) rebuie sa apatina multimii divizorilor lui 12.dar div lui 12, D12=(1,2,3,4,6,12)
DECI:n+3=1 n=-2 nu e nr natutral
         n+3=2 n=-1 nu e nr natural
         n+3=3 n=0
                n+3=4 n=1
n+3=6 n=3
n+3=12 n=9.deci n=(0,1,3,9)
deci A=(16,9,4,1)