a/ $B$ là phân số $→n-3\ne 0$
$→n\ne 3$
b/ $n=-2(tm)→B=\dfrac{4}{-2-3}=-\dfrac{4}{5}$
$n=0(tm)→B=\dfrac{4}{0-3}=-\dfrac{4}{3}$
$n=10(tm)→B=\dfrac{4}{10-3}=\dfrac{4}{7}$
c/ $B\in\mathbb Z$
$→\dfrac{4}{n-3}\in\mathbb Z$
$→4\vdots n-3$
$→n-3in Ư(4)=\{±1;±2;±4\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline n-3&1&-1&2&-2&4&-4\\\hline n&4&2&5&1&7&-1\\\hline\quad&tm&tm&tm&tm&tm&tm\\\hline\end{array}$
Vậy $n\in \{4;2;5;1;7;-1\}$