2n+3 ⋮ n+1
=> 2n+2+1 ⋮ n+1
=> 2(n+1)+1 ⋮ n+1
Vì 2(n+1) ⋮ n+1 nên để 1 ⋮ n+1
=> n+1 \(\in\) Ư(1) = {1}
+) n+1 = 1 => n=0
Vậy n = {0}
\(\frac{2n+3}{n+1}\)
\(=\frac{2n+2+1}{n+1}\)
\(=2+\frac{1}{n+1}\)
\(\Rightarrow\left(n+1\right)\inƯ\left(1\right)=\left\{\pm1\right\}\)
\(\Rightarrow n\in\left\{0;-2\right\}\)
mà n là số tự nhiên
\(\Rightarrow n=0\)