Để \(\dfrac{3}{x+2}\) nhận giá trị nguyên thì \(3⋮\left(x+2\right)\)
\(\Rightarrow:\left(x+2\right)\inƯ\left(3\right)\)
\(hay:\left(x+2\right)\in\left\{\pm1;\pm3\right\}\)
TH1: x + 2 = -1
\(x+2=-1\\ x=-1-2\\ x=-3\)
TH2: x + 2 = 1
\(x+2=1\\ x=1-2\\ x=-1\)
TH3: x + 2 = -3
\(x+2=-3\\ x=-3-2\\ x=-5\)
TH4: x + 2 = 3
\(x+2=3\\ x=3-2\\ x=1\)
Vậy \(x\in\left\{1;-1;-3;-5\right\}\)