\(\frac{10n-1}{5n+2}=\frac{10n+4-5}{5n+2}=\frac{2.\left(5n+2\right)}{5n+2}-\frac{5}{5n+2}=2-\frac{5}{5n+2}\)
Để \(\frac{10n-1}{5n+2}\) nguyên thì \(\frac{5}{5n+2}\) nguyên
Vậy 5n + 2 \(\in\) Ư(5)
\(\Leftrightarrow\) 5n + 2 \(\in\) {-5;-1;1;5}
\(\Leftrightarrow\) 5n \(\in\) {-7;-3;-1;3}
\(\Rightarrow\) n rỗng.