Đáp án:
$\begin{array}{l}
A = \dfrac{6}{{1.4}} + \dfrac{6}{{4.7}} + \dfrac{6}{{7.10}} + ... + \dfrac{6}{{n\left( {n + 3} \right)}}\\
= 2.\left( {\dfrac{3}{{1.4}} + \dfrac{3}{{4.7}} + \dfrac{3}{{7.10}} + ... + \dfrac{3}{{n\left( {n + 3} \right)}}} \right)\\
= 2.\left( {\dfrac{{4 - 1}}{{1.4}} + \dfrac{{7 - 4}}{{4.7}} + \dfrac{{10 - 7}}{{7.10}} + ... + \dfrac{{n + 3 - n}}{{n\left( {n + 3} \right)}}} \right)\\
= 2.\left( {1 - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} + ... + \dfrac{1}{n} - \dfrac{1}{{n + 3}}} \right)\\
= 2.\left( {1 - \dfrac{1}{{n + 3}}} \right)\\
= 2.\dfrac{{n + 2}}{{n + 3}}\\
Do:\dfrac{{n + 2}}{{n + 3}} < 1\\
\Leftrightarrow 2.\dfrac{{n + 2}}{{n + 3}} < 2.1\\
\Leftrightarrow A < 2\\
Vậy\,A < 2
\end{array}$
