[tex]\frac{n+1}{n+3} =\frac{n+3-2}{n+3} =1-\frac{2}{n+3}[/tex]
Ca sa fie reductibila
2 | n+3
n+3={2,4,6,8...,2024}
n={1,3,5,...,2021}
[tex]\frac{n+2}{n+5} =\frac{n+5-3}{n+5} =1-\frac{3}{n+5}[/tex]
3 | n+5
n+5={3,6,9,...,2025}
n={1,4,7,...,2020}
Ca sa fie simultan, se face intersectia solutiilor, adica solutiile comune
n={1,7,13,19,...,2017}
Adica (2017-1):6+1=336+1=337 numere
