Ta có : $P = \dfrac{2n-5}{3n-2}$
Để $P$ nguyên thì : $2n-5 \vdots 3n-2$
$\to 3.(2n-5) \vdots 3n-2$
$\to 6n-15 \vdots 3n-2$
$\to 2.(3n-2) - 11 \vdots 3n-2$
$\to 11 \vdots 3n-2$
$\to 3n-2 \in Ư(11)$
$\to 3n-2 \in \{-1,1,-11,11\}$
$\to 3n \in \{1,3,-9,13\}$
$\to n \in \{\dfrac{1}{3},1,-3,\dfrac{13}{3}\}$
Mà : $n \in Z$
$\to n \in \{-3,1\}$