Ta có:
$\dfrac{1}{\sqrt{k}+\sqrt{k+4}}=\dfrac{\sqrt{k+4}-\sqrt{k}}{4}$
Từ đó suy ra:
$\dfrac{1}{1+\sqrt5}+\dfrac{1}{\sqrt5+\sqrt9}+\dfrac{1}{\sqrt9+\sqrt{13}}+\ldots+\dfrac{1}{\sqrt{2021}+\sqrt{2025}}$
$=\dfrac{\sqrt5-1}{4}+\dfrac{\sqrt9-\sqrt5}{4}+\dfrac{\sqrt{13}-\sqrt9}{4}+\ldots+\dfrac{\sqrt{2025}-\sqrt{2021}}{4}$
$=\dfrac{\sqrt{2025}-1}{4}=11$