\(\frac{n}{n+3}\)=\(\frac{n+3-3}{n+3}\)=\(\frac{n+3}{n+3}\)-\(\frac{3}{n+3}\)=1-\(\frac{3}{n+3}\)\(\Rightarrow\)3\(⋮\)n+3\(\Rightarrow\)n+3\(\in\)Ư(3)=\(\left\{-3;-1;1;3\right\}\)
+ n+3=-3\(\Rightarrow\)n=-3-3=-6
+ n+3=-1\(\Rightarrow\)n=-1-3=-4
+ n+3=1\(\Rightarrow\)n=1-3=-2
+n+3=3\(\Rightarrow\)n=3-3=0
Với n \(\in\)(-6;-4;-2;0) thì \(\frac{n}{n+3}\)có giá trị nguyên