a/ \(\dfrac{4n+1}{2n+1}\\=\dfrac{4n+2-1}{2n+1}\\=2-\dfrac{1}{2n+1}∈\Bbb Z\\→1\vdots 2n+1\\→2n+1∈Ư(1)=\{±1\}\\→2n∈\{-2;0\}\\→n∈\{-1;0\}\)
Vậy \(n∈\{-1;0\}\)
b/ \(\dfrac{3n+4}{n-1}\\=\dfrac{3n-3+7}{n-1}\\=3+\dfrac{7}{n-1}∈\Bbb Z\\→7\vdots n-1\\→n-1∈Ư(7)=\{±1;±7\}\\→n∈\{2;0;8;-6\}\)
Vậy \(n∈\{2;0;8;-6\}\)