Hướng dẫn giảiChọn D$$A = \left\{{x \in \mathbb{R}\left|{x^2 + x + 1 = 0}\right.}\right\}$$. Ta có $$x^2 + x + 1 = 0\, \left({vn}\right)$$ $$\Rightarrow A = \emptyset$$.$$B = \left\{{x \in \mathbb{N}\left|{x^2 - 2 = 0}\right.}\right\}$$. Ta có $$x^2 - 2 = 0$$ $$\Leftrightarrow x = \pm \sqrt 2 \notin \mathbb{N}$$ $$\Rightarrow B = \emptyset$$$$C=\left\{ x\in \mathbb{Z}\left| \left( {{x}^{3}}-3 \right)\left( {{x}^{2}}+1 \right)=0 \right. \right\}$$. Ta có $$\left( {{x}^{3}}-3 \right)\left( {{x}^{2}}+1 \right)=0$$ $$\Leftrightarrow x = \sqrt[3]{3}\notin \mathbb{Z}$$ $$\Rightarrow C = \emptyset$$$$D = \left\{{x \in \mathbb{Q}\left|{x\left({x^2 + 3}\right) = 0}\right.}\right\}$$. Ta có $$x\left({x^2 + 3}\right) = 0$$ $$\Leftrightarrow x = 0$$ $$\Rightarrow D = \left\{0 \right\}.$$