Đáp án đúng: B $\displaystyle \sqrt[3]{{{n}^{3}}-2{{n}^{2}}}-n\sim \sqrt[3]{{{n}^{3}}}-n=0\xrightarrow[{}]{}$ nhân lượng liên hợp : $\displaystyle \lim \left( \sqrt[3]{{{n}^{3}}-2{{n}^{2}}}-n \right)=\lim \frac{-2{{n}^{2}}}{\sqrt[3]{{{\left( {{n}^{3}}-2{{n}^{2}} \right)}^{2}}}+n.\sqrt[3]{{{n}^{3}}-2{{n}^{2}}}+{{n}^{2}}}=\lim \frac{-2}{\sqrt[3]{{{\left( 1-\frac{2}{n} \right)}^{2}}}+\sqrt[3]{1-\frac{2}{n}}+1}=-\frac{2}{3}.$ Chọn B. Giải nhanh: $\displaystyle \sqrt[3]{{{n}^{3}}-2{{n}^{2}}}-n=\frac{-2{{n}^{2}}}{\sqrt[3]{{{\left( {{n}^{3}}-2{{n}^{2}} \right)}^{2}}}+n.\sqrt[3]{{{n}^{3}}-2{{n}^{2}}}+{{n}^{2}}}\sim \frac{-2{{n}^{2}}}{\sqrt[3]{{{n}^{6}}}+n.\sqrt[3]{{{n}^{3}}}+{{n}^{2}}}=-\frac{2}{3}.$