Question

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

Answer 1

Xét dạng tổng quát :

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}\)= \(\dfrac{\left(\sqrt{k+1}-\sqrt{k}\right)}{k+1-k}=\sqrt{k+1}-\sqrt{k}\) ( với k > 0 )

Ta có : \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

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

= \(\sqrt{100}-\sqrt{1}=10-1=9\)