Đáp án:
$\begin{array}{l}
10)\\
\lim \frac{{3n - \sqrt {4{n^3} + 9{n^2} + 16} }}{{n\sqrt {n - 2} + 2n + 1}}\\
= \lim \frac{{\frac{{3n}}{{n\sqrt n }} - \sqrt {4 + \frac{9}{n} + \frac{{16}}{{{n^3}}}} }}{{\sqrt {1 - \frac{2}{n}} + \frac{2}{{\sqrt n }} + \frac{1}{{n\sqrt n }}}}\left( {chia\,ca\,tu + mau\,cho\,n\sqrt n } \right)\\
= \lim \frac{{\frac{3}{{\sqrt n }} - \sqrt {4 + \frac{9}{n} + \frac{{16}}{{{n^3}}}} }}{{\sqrt {1 - \frac{2}{n}} + \frac{2}{{\sqrt n }} + \frac{1}{{n\sqrt n }}}}\\
= \frac{{ - \sqrt 4 }}{1}\\
= - 2\\
\Rightarrow B\\
11)\\
\lim \frac{{\sqrt {8n + 9} - \sqrt[4]{{4{n^2} - 9n + 25}}}}{{\sqrt[6]{{8{n^3} + 16{n^2} + 15}} + \sqrt {2n + 5} }}\\
= \lim \frac{{\frac{{\sqrt {8n + 9} }}{{\sqrt n }} - \frac{{\sqrt[4]{{4{n^2} - 9n + 25}}}}{n}}}{{\frac{{\sqrt[6]{{8{n^3} + 16{n^2} + 15}}}}{{\sqrt n }} + \frac{{\sqrt {2n + 5} }}{{\sqrt n }}}}\\
= \lim \frac{{\sqrt {8 + \frac{9}{n}} - \sqrt[4]{{4 - \frac{9}{n} + \frac{{25}}{{{n^2}}}}}}}{{\sqrt[6]{{8 + \frac{{16}}{n} + \frac{{15}}{{{n^3}}}}} + \sqrt {2 + \frac{5}{n}} }}\\
= \frac{{\sqrt 8 - \sqrt 2 }}{{\sqrt[6]{8} + \sqrt 2 }}\\
= \frac{{2\sqrt 2 - \sqrt 2 }}{{\sqrt 2 + \sqrt 2 }}\\
= \frac{1}{2}
\end{array}$
