Lời giải:
$2n-1$ là ước của $3n+4$
Ta có:
$(3n+4) \vdots (2n-1)$
$=>2.(3n+4) \vdots (2n-1)$
$=>(6n+8) \vdots (2n-1)$
$=>(6n-3+11) \vdots (2n-1)$
$=>3.(2n-1)+11 \vdots (2n-1)$
Mà $3.(2n-1) \vdots (2n-1)$
$=>11 \vdots (2n-1)$
$=>2n-1∈U(11)=${$-11;-1;1;11$}
$2n-1=-11<=>2n=-10<=>n=-5$
$2n-1=-1<=>2n=0<=>n=0$
$2n-1=1<=>2n=2<=>n=1$
$2n-1=11<=>2n=12<=>n=6$
Vậy $n=${$-5;0;1;6$}
