Giải thích các bước giải:
$\begin{array}{l}
\lim \left( {\sqrt[3]{{{n^2} - 27{n^3}}} + \sqrt {9{n^2} + 7n} } \right)\\
= \lim \left( {\sqrt[3]{{{n^2} - 27{n^3}}} + 3n + \sqrt {9{n^2} + 7n} - 3n} \right)\\
= \lim \left( {\dfrac{{{n^2} - 27{n^3} + 27{n^3}}}{{{{\left( {3n} \right)}^2} - 3n.\sqrt[3]{{{n^2} - 27{n^3}}} + {{\left( {\sqrt[3]{{{n^2} - 27{n^3}}}} \right)}^2}}} + \dfrac{{9{n^2} + 7n - 9{n^2}}}{{\sqrt {9{n^2} + 7n} + 3n}}} \right)\\
= \lim \left( {\dfrac{{{n^2}}}{{9{n^2} - 3n.\sqrt[3]{{{n^2} - 27{n^3}}} + \left( {\sqrt[3]{{{n^2} - 27{n^3}}}} \right)}} + \dfrac{{7n}}{{\sqrt {9{n^2} + 7n} + 3n}}} \right)\\
= \lim \left( {\dfrac{1}{{9 - 3\sqrt[3]{{\dfrac{1}{n} - 27}} + {{\left( {\sqrt[3]{{\dfrac{1}{n} - 27}}} \right)}^2}}} + \dfrac{7}{{\sqrt {9 + \dfrac{7}{n}} + 3}}} \right)\\
= \dfrac{1}{{9 - 3\sqrt[3]{{ - 27}} + {{\left( {\sqrt[3]{{ - 27}}} \right)}^2}}} + \dfrac{7}{{\sqrt 9 + 3}}\\
= \dfrac{{65}}{{54}}
\end{array}$
