$\dfrac{2n-1}{n^2-1} \in Z$
$\to 2n-1 \vdots n^2-1$
$\to 2n-1 \vdots (n-1)(n+1)$
$\to \begin{cases}2n-1 \vdots n-1\\2n-1 \vdots n+1\\\end{cases}$
$\to \begin{cases}2n-2+1 \vdots n-1\\2n+2 -3 \vdots n+1\\\end{cases}$
$\to \begin{cases}1 \vdots n-1\\3 \vdots n+1\\\end{cases}$
$\to \begin{cases}n-1 \in Ư(1)={+-1}\\n+1 \in Ư(3)={+-1,+-3}\\\end{cases}$
$\to \begin{cases}\left[ \begin{array}{l}n=0\\n=2\end{array} \right.\\\left[ \begin{array}{l}n=0\\n=-2\\n=2\\n=-4\end{array} \right.\\\end{cases}$
$\to \left[ \begin{array}{l}n=2\\n=0\end{array} \right.$
Vậy với $\left[ \begin{array}{l}n=2\\n=0\end{array} \right.$ thì $\dfrac{2n-1}{n^2-1} \in Z$
