Để $\dfrac{n+3}{2n-5}$ $∈$ $Z$ thì : $n+3 \vdots 2n-5$
$⇔ 2.(n+3) - (2n-5) \vdots 2n-5$
$⇔ 2n + 6 - 2x + 5 \vdots 2n-5$
$⇔ 11 \vdots 2n-5$
$⇒ 2n-5$ $∈$ `Ư(11)={±1;±11}`
$⇔2n$ $∈$ `{-6;4;6;16}`
$⇔n$ $∈$ `{-3;2;3;8}`
Vậy $n$ $∈$ `{-3;2;3;8}` thì $\dfrac{n+3}{2n-5}$ $∈$ $Z$.