Ta có:\(\dfrac{x+3}{x-2}\)=\(\dfrac{x-2+5}{x-2}\)=1+\(\dfrac{5}{x-2}\)
Để \(\dfrac{x+3}{x-2}\) là số nguyên thì \(\dfrac{5}{x-2}\) phải là số nguyên
=>x-2∈Ư(5)=\(\left\{{}\left\{1;-1;5;-5\right\}}\)
+)x-2=1 +)x-2=-1 +)x-2=5 +)x-2=-5
=>x=3 =>x=1 x=7 x=-3
Vậy để\(\dfrac{x+3}{x-2}\) thì x∈\(\left\{3;7;1;-3\right\}\)