*Lời giải :
`B = (2n - 3)/(n + 1)`
`⇔ 2n - 3 \vdots n + 1`
`⇔ 2 (n - 1) - 5 \vdots n + 1`
`⇔ 5 \vdots n + 1`
`⇔ n + 1 ∈ Ư (5) = {±1; ±5}`
Vì `x ∈ ZZ`
`⇔` \(\left\{ \begin{array}{l}n+1=1\\n+1=-1\\n+1=5\\n+1=-5\end{array} \right.\)
`⇔` \(\left\{ \begin{array}{l}n=0\\n=-2\\n=4\\n=-5\end{array} \right.\)