Question

Va rog sa ma ajutați cu exercițiul de mai jos și sa-mi explicați cum se rezolva ! A2 ma refer ... mulțumesc !

Answer 1

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

$a=\frac{\sqrt{1}-\sqrt{2}}{(\sqrt{1}+\sqrt{2})(\sqrt{1}-\sqrt{2})}+\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})}+...+\frac{\sqrt{98}-\sqrt{99}}{(\sqrt{98}+\sqrt{99})(\sqrt{98}-\sqrt{99})}$

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

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

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

$a=\frac{\sqrt{1}-\sqrt{99}}{-1}$

$a=\frac{-(\sqrt{99}-\sqrt{1})}{-1}$

$a=\frac{\sqrt{99}-\sqrt{1}}{1}$

$a=\sqrt{99}-\sqrt{1}$

$a=\sqrt{3^{2}*11}-1$

$\boxed{a=3\sqrt{11}-1}$