Question

$\frac{1}{1+\sqrt{2} } +\frac{1}{\sqrt{2}+\sqrt{3}}+ .......+\frac{1}{\sqrt{98}+\sqrt{99}} =?$

Answer 1

$\frac{1}{1+\sqrt{2} }=\frac{1*(1-\sqrt{2}) }{(1+\sqrt{2})(1-\sqrt{2})}=\frac{1-\sqrt{2} }{-1}=\sqrt{2}-1\\\\\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{1*(\sqrt{2}-\sqrt{3})} {(\sqrt{2} +\sqrt{3})(\sqrt{2}-\sqrt{3})}=\frac{\sqrt{2}-\sqrt{3}}{-1}=\sqrt{3}-\sqrt{2}$

И так далее

$\frac{1}{\sqrt{98}+\sqrt{99}}=\frac{1*(\sqrt{98}-\sqrt{99})}{(\sqrt{98}+\sqrt{99})(\sqrt{98}-\sqrt{99})}=\frac{\sqrt{98}-\sqrt{99}}{-1} =\sqrt{99}-\sqrt{98}$

Получим :

$\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{99}-\sqrt{98}=-1+\sqrt{99}=3\sqrt{11}-1$