Ta có:\(\dfrac{1}{\sqrt{n}+\sqrt{n+1}}=\dfrac{\sqrt{n+1}^2-\sqrt{n}^2}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}\)

Do đó:

\(\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1;\dfrac{1}{\sqrt{2}+\sqrt{3}}=\sqrt{3}-\sqrt{2};...;\dfrac{1}{\sqrt{99}-\sqrt{100}}=\sqrt{100}-\sqrt{99}\)

Đến đây bạn tự giải tiếp nhé.