$A=\frac{n+1}{n-3}=$ $\frac{(n-3)+4}{n-3}=$ $1+\frac{4}{n-3}$
⇒n-3∈Ư(4)={1;2;4}
n-3=1⇒n=4
n-3=2⇒n=5
n-3=4⇒n=7
Vậy n∈{4;5;7}
$B=\frac{n+1}{n-2}=$ $\frac{(n-2)+3}{n-2}=$ $1+\frac{3}{n-2}$
⇒n-2∈Ư(3)={1;3}
n-2=1⇒n=3
n-2=3⇒n=5
Vậy n∈{3;5}
$C=\frac{10n}{5n-3}=$ $\frac{2(5n-3)+6}{5n-3}=$ $2+\frac{6}{5n-3}$
⇒5n-3∈Ư(6)={1;2;3;6}
5n-3=1⇒n=4/5 (loại)
5n-3=2⇒n=1
5n-3=3⇒n=6/5 (loại)
5n-3=6⇒n=9/5 (loại)
Vậy n∈{1}