Đáp án:
$\begin{array}{l}
\lim \left( {\sqrt[3]{{{n^3} - 2{n^2}}} - n} \right)\\
= \lim \frac{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}} - n} \right)\left( {\sqrt[3]{{{{\left( {{n^3} - 2{n^2}} \right)}^2}}} + n\sqrt[3]{{{n^3} - 2{n^2}}} + {n^2}} \right)}}{{{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}}} \right)}^2} + n.\sqrt[3]{{{n^3} - 2{n^2}}} + {n^2}}}\\
= \lim \frac{{ - 2{n^2}}}{{{{\left( {\sqrt[3]{{{n^3} - 2{n^2}}}} \right)}^2} + n.\sqrt[3]{{{n^3} - 2{n^2}}} + {n^2}}}\\
= \lim \frac{{ - 2}}{{{{\left( {\sqrt[3]{{1 - \frac{2}{n}}}} \right)}^2} + \sqrt[3]{{1 - \frac{2}{n}}} + 1}}\\
= \frac{{ - 2}}{{1 + 1 + 1}} = - \frac{2}{3}
\end{array}$
( Chia cả tử và mẫu cho x^2)
