\(B=\dfrac{n+3-5}{n+3}=1+\dfrac{-5}{n+3}\)
Để B thuộc Z thì \(\left(n+3\right)\inƯ\left(5\right)=\left\{-1;1;-5;5\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}n+3=-1\\n+3=1\\n+3=5\\n+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}n=-4\\n=-2\\n=2\\n=-8\end{matrix}\right.\)