Question

Tính:  \(B=\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+....+\frac{1}{\sqrt{97}+\sqrt{99}}\)

Answer 1

\(B=\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+...+\frac{1}{\sqrt{97}+\sqrt{99}}.\)

\(=\frac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}+...+\frac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{99}+\sqrt{97}\right)\left(\sqrt{99}-\sqrt{97}\right)}.\)

\(=\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+...+\frac{\sqrt{99}-\sqrt{97}}{99-97}.\)

\(=\frac{\sqrt{5}}{2}-\frac{\sqrt{3}}{2}+\frac{\sqrt{7}}{2}-\frac{\sqrt{5}}{2}+...+\frac{\sqrt{99}}{2}-\frac{\sqrt{97}}{2}=\frac{\sqrt{99}}{2}-\frac{\sqrt{3}}{2}\)

Vậy \(B=\frac{\sqrt{99}-\sqrt{3}}{2}.\)