Ta có
$\sqrt{1} + \sqrt{2} > \sqrt{1}$
nên
$\dfrac{1}{\sqrt{1} + \sqrt{2}} < \dfrac{1}{\sqrt{1}}$
Tương tự ta có
$\dfrac{1}{\sqrt{2} + \sqrt{3}} < \dfrac{1}{\sqrt{2}}$
$\dfrac{1}{\sqrt{3} + \sqrt{4}} < \dfrac{1}{\sqrt{3}}$
...
$\dfrac{1}{\sqrt{24} + \sqrt{25}} < \dfrac{1}{\sqrt{24}}$
Cộng từng vế ta có
$\dfrac{1}{\sqrt{1} + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + \dfrac{1}{\sqrt{3} + \sqrt{4}} + \cdots + \dfrac{1}{\sqrt{24} + \sqrt{25}} < \dfrac{1}{\sqrt{1}} + \dfrac{1}{\sqrt{2}} + \cdots + \dfrac{1}{\sqrt{24}}$
Vậy $A < B$.