\(\dfrac{2n+3}{n+2}=\dfrac{2\left(n-2\right)+7}{n-2}=2+\dfrac{7}{n-2}\)
=> Để 2n + 3 chia hết cho n - 2
thì \(n-2\inƯ\left(7\right)\)
\(\Leftrightarrow n-2=\left\{-7;-1;1;7\right\}\)
<=> \(n=\left\{-5;1;3;9\right\}\)
vậy..............................................