Question

Vreau a de la 20 e urgent va implor dau 100 puncte clasa a 7-a !!!!!!!!!!!!!

Answer 1

Răspuns:

1

Explicație pas cu pas:

$\sqrt{ \dfrac{102}{99} \cdot \Big[ \Big( \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + ... + \dfrac{1}{50 \cdot 51}\Big) - \Big( \dfrac{1}{51 \cdot 52} + \dfrac{1}{52 \cdot 53} + \dfrac{1}{53 \cdot 54} + ... + \dfrac{1}{101 \cdot 102}\Big)\Big]} =$

$= \sqrt{\dfrac{102}{99} \cdot \Big[ \Big( \dfrac{2-1}{1 \cdot 2} + \dfrac{3-2}{2 \cdot 3} + \dfrac{4-3}{3 \cdot 4} + ... + \dfrac{51-50}{50 \cdot 51}\Big) - \Big( \dfrac{52-51}{51 \cdot 52} + \dfrac{53-52}{52 \cdot 53} + \dfrac{54-53}{53 \cdot 54} + ... + \dfrac{102-101}{101 \cdot 102}\Big)\Big]}$

$= \sqrt{\dfrac{102}{99} \cdot \Big[ \Big( \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ... + \dfrac{1}{50} - \dfrac{1}{51}\Big) - \Big( \dfrac{1}{51} - \dfrac{1}{52} + \dfrac{1}{52} - \dfrac{1}{53} + \dfrac{1}{53} - \dfrac{1}{54} + ... + \dfrac{1}{101} - \dfrac{1}{102}\Big)\Big]}$

$= \sqrt{\dfrac{102}{99} \cdot \Big[ \Big( \dfrac{^{51)}1}{1} - \dfrac{1}{51}\Big) - \Big( \dfrac{^{2)} 1}{51} - \dfrac{1}{102}\Big)\Big]}$

$= \sqrt{\dfrac{102}{99} \cdot \Big(\dfrac{51 - 1}{51} - \dfrac{2 - 1}{102}\Big)}$

$= \sqrt{\dfrac{102}{99} \cdot \Big(\dfrac{^{2)} 50}{51} - \dfrac{1}{102}\Big)} = \sqrt{\dfrac{102}{99} \cdot \dfrac{100 - 1}{102}}$

$= \sqrt{\dfrac{102}{99} \cdot \dfrac{99}{102}} = \sqrt{1} = \bf 1$