\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(=\dfrac{\sqrt{2}-1}{\left(1+\sqrt{2}\right).\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right).\left(\sqrt{3}-2\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{99}+\sqrt{100}\right)\left(\sqrt{100}-\sqrt{99}\right)}\)\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-1=10-1=9\)
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.
