Question

Aratati ca numarul a este rational,unde :

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

Answer 1

vezi calculul in atasament

Answer 2

$a=\frac{\sqrt{1}-\sqrt{2}}{\sqrt{1}\cdot\sqrt{2}}+\frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}\cdot\sqrt{3}}+...+\frac{\sqrt{99}-\sqrt{100}}{\sqrt{99}\cdot\sqrt{100}}$
$=\frac{\sqrt{1}}{\sqrt{1}\cdot\sqrt{2}}-\frac{\sqrt{2}}{\sqrt{1}\cdot\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{2}\cdot\sqrt{3}}-\frac{\sqrt{3}}{\sqrt{2}\cdot\sqrt{3}}+...+\frac{\sqrt{99}}{\sqrt{99}\cdot\sqrt{100}}-\frac{\sqrt{100}}{\sqrt{99}\cdot\sqrt{100}}$
$=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{100}}-\frac{1}{\sqrt{99}}$
$=-\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-...-\frac{1}{\sqrt{99}}+\frac{1}{\sqrt{100}}$
$=-\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{100}}$
$=-\frac{1}{1}+\frac{1}{10}$
$=-\frac{10}{10}+\frac{1}{10}$
$=-\frac{9}{10}\in\mathbb{Q}$