[tex]\frac{2n+7}{n+3}\;este\;ireductibil\;daca\; (2n+7\;;\;n+3)=1\;![/tex]
presupunem ca exista d≠1 astfel ca d | 2n+7 si d | n+3
⇵
d | 2(n+3)=2n+6
inseamna ca d | 2n+7 - (2n+6) adica; d | 2n+7-n-6
deci d | 1
=> d =1 ceea ce inseamna ca (2n+7 ; n+3) = 1
adica; [tex]\;\;\frac{2n+7}{n+3}\;este\;ireductibil\;[/tex]