$mx^{2}-(2m-3)x+m+2=0$
$Δ=(2m-3)^{2}-4.m.(m+2)$
$=4m^{2}-12m+9-4m^{2}-8m$
$=-20m+9$
$⇒x_{1}=\dfrac{2m-3+\sqrt{-20m+9}}{2m}$
$x_{2}=\dfrac{2m-3+\sqrt{-20m+9}}{2m}$
Vì $x_{1}$ và $x_{2}∈Z$
$⇒x_{1}+x_{2}∈Z$
$⇒\dfrac{2m-3+\sqrt{-20m+9}}{2m}+\dfrac{2m-3-\sqrt{-20m+9}}{2m}$
$=\dfrac{2m-3}{m}=2-\dfrac{3}{m}$
$⇒m∈Ư_{3}=\{-3;-1;1;3\}$
Vì $m>-3⇒m∈\{±1;3\}$