Question

Aratati ca numarul a este rational,unde
a=√1-√2/√1×2+√2-√3/√3×4+...+√99-√100/√99×100

Answer 1

$A= \frac{ \sqrt{1}- \sqrt{2} }{ \sqrt{1 \cdot 2} } + \frac{ \sqrt{2}- \sqrt{3} }{ \sqrt{2 \cdot 3} }+...+ \frac{ \sqrt{99}- \sqrt{100} }{ \sqrt{99 \cdot 100} } = \\ \\ ~~~= \frac{ \sqrt{1} }{ \sqrt{1} \cdot \sqrt{2} }- \frac{ \sqrt{2} }{ \sqrt{1} \cdot \sqrt{2} } + \frac{ \sqrt{2} }{ \sqrt{2} \cdot \sqrt{3} } - \frac{ \sqrt{3} }{ \sqrt{2} \cdot \sqrt{3} } +...+ \frac{\sqrt{99}}{\sqrt{99} \cdot \sqrt{100}}- \frac{\sqrt{100}}{\sqrt{99} \cdot \sqrt{100}} =$

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