\(\dfrac{x^2+2}{x+2}=\dfrac{x^2+2x-2x-4+6}{x+2}=x-2+\dfrac{6}{x+2}\) nguyên
\(\Rightarrow6⋮x+2\)
\(\Rightarrow x+2\in U\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Rightarrow x\in\left\{-1;-3;0;-4;1;-5;4;-8\right\}\)
\(A=\dfrac{x^2+2}{x+2}=\dfrac{\left(x^2-4\right)+6}{x+2}=x-2+\dfrac{6}{x+2}\) \(\left(x\ne-2\right)\)
\(A\in Z\Leftrightarrow\dfrac{6}{x+2}\in Z\)
\(\Leftrightarrow x+2\in U\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3;0;-4;1;-5;4;-8\right\}\)
Vậy \(x\in\left\{-8;-5;-4;-3;-1;0;1;4\right\}\) thỏa đề bài
