Đáp án:
$\begin{array}{l}
Do:{2^2} < 2.3\\
\Rightarrow \frac{1}{{{2^2}}} > \frac{1}{{2.3}}\\
TT:\frac{1}{{{3^2}}} > \frac{1}{{3.4}};\frac{1}{{{4^2}}} > \frac{1}{{4.5}};...\frac{1}{{{{10}^2}}} > \frac{1}{{10.11}}\\
\Rightarrow \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{10}^2}}}\\
> \frac{1}{{2.3}} + \frac{1}{{3.4}} + \frac{1}{{4.5}} + ... + \frac{1}{{10.11}}\\
\Rightarrow S > \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} + ... + \frac{1}{{10}} - \frac{1}{{11}}\\
\Rightarrow S > \frac{1}{2} - \frac{1}{{11}}\\
\Rightarrow S > \frac{{11 - 2}}{{22}}\\
\Rightarrow S > \frac{9}{{22}}
\end{array}$
