Để x là số nguyên thì: \(2n+5⋮n+3\)
\(\Leftrightarrow2n+6-1⋮n+3\)
\(\Leftrightarrow2\left(n+3\right)-1⋮n+3\)
Do 2(n+3) \(⋮\) n+3 \(\Rightarrow1⋮2n+3\) \(\Rightarrow2n+3\inƯ\left(1\right)=\left\{1;-1\right\}\)
\(\Rightarrow2n\in\left\{-2;-4\right\}\)
\(\Rightarrow n\in\left\{-1;-2\right\}\)
Vậy \(n\in\left\{-1;-2\right\}\)
\(\frac{2n+5}{n+3}=\frac{n+3+n+3-1}{n+3}\)
\(=\frac{n+3}{n+3}+\frac{n+3}{n+3}-\frac{1}{n+3}\)
=1+1-\(\frac{1}{n+3}\)
=2-\(\frac{1}{n+3}\)
Để \(\frac{2n+5}{n+3}\)là số nguyên thì \(\frac{1}{n+3}\) phải là số nguyên
Suy ra: n+3∈U(1)={1;-1}
n+3=1=>n=-2
n+3=-1=>n=-4
Vậy n=-2 hoặc n=-4