$\dfrac{6n+5}{2n-1}∈\Bbb Z(n\ne \dfrac{1}{2},n∈\Bbb Z)\\→\dfrac{6n-3+8}{2n-1}∈\Bbb Z\\→\dfrac{3(2n-1)+8}{2n-1}∈\Bbb Z\\→3+\dfrac{8}{2n-1}∈\Bbb Z\\→8\vdots 2n-1\\→2n-1∈Ư(8)=\{±1;±2;±4;±8\}$
$→$ Ta có bảng:
$\begin{array}{|c|c|c|}\hline 2n-1&1&-1&2&-2&4&-4&8&-8\\\hline 2n&2&0&3&-1&5&-3&9&-7\\\hline n&1&0&\dfrac{3}{2}&-\dfrac{1}{2}&\dfrac{5}{2}&-\dfrac{3}{2}&\dfrac{9}{2}&-\dfrac{7}{2}\\\hline \quad&tm&tm&ktm&ktm&ktm&ktm&ktm&ktm\\\hline\end{array}$
Vậy $n∈\left\{1;0\right\}$
