Ta có : n chia hết cho 3 + n
Suy ra : n chia hết cho 3
=> n thuộc B(3) = {....................-3;-6;-9;0;3;6;9;................}
ta có : \(\frac{n}{3+n}=\frac{\left(3+n\right)-3}{3+n}\)
vì \(\left(3+n\right)⋮\left(3+n\right)\)để \(\frac{\left(3+n\right)-3}{3+n}\)nguyên \(\Leftrightarrow-3⋮\left(3+n\right)\Leftrightarrow\left(3+n\right)\inƯ\left(-3\right)\)
\(\RightarrowƯ\left(-3\right)=1;-1;3;-3\)
\(\Rightarrow3+n=1\Rightarrow n=-2\)
\(\Rightarrow3+n=-1\Rightarrow n=-4\)
\(\Rightarrow3+n=3\Rightarrow n=0\)
\(\Rightarrow3+n=-3\Rightarrow n=-6\)