Question

C/minh: \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=9\)

Answer 1

Dạng tổng quát :

\(\frac{1}{\sqrt{a-1}+\sqrt{a}}=\frac{\sqrt{a-1}-\sqrt{a}}{\left(\sqrt{a-1}+\sqrt{a}\right)\left(\sqrt{a-1}-\sqrt{a}\right)}\)

\(=\frac{\sqrt{a-1}-\sqrt{a}}{a-1-a}=\sqrt{a}-\sqrt{a-1}\)

Do đó :

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

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

\(=10-1=9\)