Đáp án:
$\begin{array}{l}
A = \dfrac{1}{{{4^2}}} + \dfrac{1}{{{6^2}}} + \dfrac{1}{{{8^2}}} + \dfrac{1}{{{{10}^2}}} + .. + \dfrac{1}{{{{160}^2}}}\\
= \dfrac{1}{{{2^2}}}\left( {\dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{{{80}^2}}}} \right)\\
= \dfrac{1}{{{2^2}}}.B\\
Do:B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{{{80}^2}}}\\
> \dfrac{1}{{{2^2}}} + \left( {\dfrac{1}{{3.4}} + \dfrac{1}{{4.5}} + ... + \dfrac{1}{{80.81}}} \right)\\
> \dfrac{1}{4} + \left( {\dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{{80}} - \dfrac{1}{{81}}} \right)\\
> \dfrac{1}{4} + \left( {\dfrac{1}{3} - \dfrac{1}{{81}}} \right)\\
> \dfrac{1}{4} + \dfrac{{26}}{{81}} > \dfrac{1}{2}\\
\Leftrightarrow B > \dfrac{1}{2}\\
\Leftrightarrow A > \dfrac{1}{{{2^2}}}.\dfrac{1}{2}\\
\Leftrightarrow A > \dfrac{1}{8}\\
Lai\,co:\\
B = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{{{80}^2}}}\\
< \dfrac{1}{4} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{79.80}}\\
\Leftrightarrow B < \dfrac{1}{4} + \left( {\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{79}} - \dfrac{1}{{80}}} \right)\\
\Leftrightarrow B < \dfrac{1}{4} + \dfrac{1}{2} - \dfrac{1}{{80}}\\
\Leftrightarrow B < \dfrac{3}{4} - \dfrac{1}{{80}} < \dfrac{3}{4}\\
\Leftrightarrow A < \dfrac{1}{{{2^2}}}.B\\
\Leftrightarrow A < \dfrac{1}{4}.\dfrac{3}{4}\\
\Leftrightarrow A < \dfrac{3}{{16}}\\
Vậy\,\dfrac{1}{8} < A < \dfrac{3}{{16}}
\end{array}$
