Folosim formula:
[tex]\it \dfrac{1}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}[/tex]
Suma din prima paranteză devine:
[tex]\it \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\ ...\ +\dfrac{1}{24}-\dfrac{1}{25}=\ 1-\dfrac{1}{25}=\dfrac{24}{25}[/tex]
Suma din a doua paranteză devine:
[tex]\it \dfrac{1}{25}-\dfrac{1}{26}+\dfrac{1}{26}-\dfrac{1}{27}+\ ...\ +\dfrac{1}{49}-\dfrac{1}{50}=\dfrac{^{2)}1}{\ \ 25}-\dfrac{1}{50}=\dfrac{1}{50}[/tex]
Acum, paranteza dreaptă devine:
[tex]\it \dfrac{^{2)}24}{\ 25}-\dfrac{1}{50}=\dfrac{48-1}{50}=\dfrac{47}{50}[/tex]
Vom calcula numărul x:
[tex]\it\ x=\sqrt{\dfrac{1}{94}\cdot\dfrac{47}{50}}=\sqrt{\dfrac{\ \ 47^{(47}}{50\cdot94}}=\sqrt{\dfrac{1}{50\cdot2}}=\sqrt{\dfrac{1}{100}}=\dfrac{1}{10}\in\mathbb{Q}[/tex]
