a, Để $B$ là 1 phân số thì: $n+2\neq0$
⇔ $n\neq-2$
b, $|B|=\frac{3}{2}$
⇔ \(\left[ \begin{array}{l}B=\frac{3}{2}\\B=\frac{-3}{2}\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}\frac{n-3}{n+2}=\frac{3}{2}\\\frac{n-3}{n+2}=\frac{-3}{2}\end{array} \right.\)
Với $\frac{n-3}{n+2}=\frac{3}{2}$
⇔ $2.(n-3)=3.(n+2)$
⇔ $2n-6=3n+6$
⇔ $n=-12$ tm
Với $\frac{n-3}{n+2}=\frac{-3}{2}$
⇔ $2.(n-3)=-3.(n+2)$
⇔ $2n-6=-3n-6$
⇔ $n=0$ tm
c, Để $B∈Z$ thì $n-3$⋮ $n+2$
⇔ $n+2-5$⋮ $n+2$
⇒ $n+2∈Ư(5)=\{-5;-1;1;5\}$
⇔ $n∈\{-7;-3;-1;3\}$ tm