Question

Помогите пожалуйста с решением #704 (7)б

Answer 1

$\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}=\\\\ =\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{100}+\sqrt{99}}=\\\\ =\frac{\sqrt{2}-1}{(\sqrt{2}+1)*(\sqrt{2}-1)}+\frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}+\frac{\sqrt{4}-\sqrt{3}}{(\sqrt{4}+\sqrt{3})(\sqrt{4}-\sqrt{3})}+...\\...+\frac{\sqrt{100}-\sqrt{99}}{(\sqrt{100}+\sqrt{99})(\sqrt{100}-\sqrt{99})}=$

$=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+\frac{\sqrt{4}-\sqrt{3}}{4-3}+...+\frac{\sqrt{100}-\sqrt{99}}{100-99}=\\\\ =\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{99}-\sqrt{98}+\sqrt{100}-\sqrt{99}=\\\\ =-1+\sqrt{100}=-1+10=9$