Ta có: $\dfrac{1}{2^2} < \dfrac{1}{1.2}$
$\dfrac{1}{3^2} < \dfrac{1}{2.3}$
$\dfrac{1}{4^2} < \dfrac{1}{3.4}$
$................................................................$
$\dfrac{1}{10^2} < \dfrac{1}{9.10}$
$⇒ A= \dfrac{1}{2^2} + \dfrac{1}{3^2} + \dfrac{1}{4^2} + ......+ \dfrac{1}{10^2} < \dfrac{1}{1.2} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + .... + \dfrac{1}{9.10} = 1 - \dfrac{1}{10} < 1$
Vậy $A<1$ ($đpcm$)