Đáp án:
$\begin{array}{l}
{2^2} > 1.2 \Rightarrow \dfrac{1}{{{2^2}}} < \dfrac{1}{{1.2}}\\
\Rightarrow \dfrac{1}{{{3^2}}} < \dfrac{1}{{2.3}};\dfrac{1}{{{4^2}}} < \dfrac{1}{{3.4}};...;\dfrac{1}{{{{150}^2}}} < \dfrac{1}{{149.150}}\\
\Rightarrow \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + .. + \dfrac{1}{{{{150}^2}}} < \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + .. + \dfrac{1}{{149.150}}\\
\Rightarrow A < 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{{149}} - \dfrac{1}{{150}}\\
\Rightarrow A < 1 - \dfrac{1}{{150}} < 1\\
Vậy\,A < 1
\end{array}$
