Question

Посчитать $\frac{1}{1+\sqrt{2} } + \frac{1}{\sqrt{2}+\sqrt{3}} + \frac{1}{\sqrt{3}+\sqrt{4}} + ... + \frac{1}{\sqrt{80}+\sqrt{81}}$

Answer 1

Ответ:

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Answer 2

$\frac{1}{1+\sqrt{2}} +\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{80}+\sqrt{81}}=\frac{1*(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}+\frac{1*(\sqrt{2}-\sqrt{3})  }{(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})} +\frac{1*(\sqrt{3}-\sqrt{4})}{(\sqrt{3}+\sqrt{4})(\sqrt{3}-\sqrt{4})}+...+\frac{1*(\sqrt{80}-\sqrt{81})}{(\sqrt{80}+\sqrt{81})(\sqrt{80}-\sqrt{81})}=\frac{1-\sqrt{2}}{1^{2}-(\sqrt{2})^{2}}+\frac{\sqrt{2}-\sqrt{3}}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}+$$\frac{\sqrt{3}-\sqrt{4}}{(\sqrt{3})^{2}-(\sqrt{4})^{2}}+...+\frac{\sqrt{80}-\sqrt{81}}{(\sqrt{80})^{2}-(\sqrt{81})^{2}}=\frac{1-\sqrt{2}}{1-2}+\frac{\sqrt{2}-\sqrt{3}}{2-3}+\frac{\sqrt{3}-\sqrt{4}}{3-4}+...+\frac{\sqrt{80}-\sqrt{81}}{80-81}=\frac{1-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} +\frac{\sqrt{3}-\sqrt{4}}{-1}+...+\frac{\sqrt{80}-\sqrt{81}}{-1}=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}...-\sqrt{80}+\sqrt{81}=-1+\sqrt{81} =-1+9=8\\\\Otvet:\boxed{8}$