Đáp án đúng: C
Giải chi tiết:Vì \({x^2} + 2\) là bội của \(x + 2\) nên \({x^2} + 2\,\, \vdots \,\,x + 2\).
\(\left. \begin{array}{l}{x^2} + 2\,\, \vdots \,\,x + 2\\x + 2\,\, \vdots \,\,x + 2\end{array} \right\} \Rightarrow \left. \begin{array}{l}{x^2} + 2x - 2x + 2\,\, \vdots \,\,x + 2\\x + 2\,\, \vdots \,\,x + 2\end{array} \right\}\)
\( \Rightarrow \left. \begin{array}{l}x\left( {x + 2} \right) - 2x + 2\,\, \vdots \,\,x + 2\\x + 2\,\, \vdots \,\,x + 2\end{array} \right\} \Rightarrow \left. \begin{array}{l} - 2x + 2\,\, \vdots \,\,x + 2\\x + 2\,\, \vdots \,\,x + 2\end{array} \right\} \Rightarrow \left\{ \begin{array}{l} - 2x + 2\,\, \vdots \,\,x + 2\\ - 2x - 4\,\, \vdots \,\,x + 2\end{array} \right. \Rightarrow 6\,\, \vdots \,\,x + 2\)
\( \Rightarrow x + 2 \in U\left( 6 \right) = \,\left\{ { \pm 1;\,\, \pm 2;\,\, \pm 3;\,\, \pm 6} \right\}\)
\( \Rightarrow x + 2 \in \left\{ { - 6;\,\, - 3;\,\, - 2;\,\, - 1;\,\,1;\,\,2;\,\,3;\,\,4} \right\}\)
\( \Rightarrow x \in \left\{ { - 8;\,\, - 5;\,\, - 4;\,\, - 3;\, - 1;\,\,0;\,\,1;\,\,2} \right\}\)
Chọn C.