\(\frac{x+2}{x^2+4}\in Z\Rightarrow x+2⋮x^2+4\)
\(\Rightarrow\left(x+2\right)\left(x-2\right)⋮x^2+4\)
\(\Rightarrow x^2-4⋮x^2+4\)
Mà \(x^2+4⋮x^2+4\)
\(\Rightarrow\left(x^2+4\right)-\left(x^2-4\right)⋮x^2+4\)
\(\Rightarrow8⋮x^2+4\)
\(\Rightarrow x^2+4\inƯ\left(8\right)\)
Mà \(x^2+4\ge0+4=4\Rightarrow x^2+4\in\left\{4;8\right\}\)
\(\Rightarrow x^2\in\left\{0;4\right\}\)
\(\Rightarrow x\in\left\{-2;0;2\right\}\)
Với \(x=-2\Rightarrow\frac{x+2}{x^2+4}=\frac{0}{4+4}=0\in Z\left(TM\right)\)
Với \(x=0\Rightarrow\frac{x+2}{x^2+4}=\frac{2}{0+4}=\frac{1}{2}\notin Z\left(0TM\right)\)
Với \(x=2\Rightarrow\frac{x+2}{x^2+4}=\frac{4}{4+4}=\frac{1}{2}\notin Z\left(0TM\right)\)
Do đó \(x=-2\)
Vậy ...
