Question

2,3 subiectul 2 va roog mult mult de tot dau coroana si 30 de puncte

Answer 1

Răspuns:

3).A.6

Explicație pas cu pas:

$\sqrt{\bigg[ \bigg(\dfrac{5}{4} + \dfrac{6}{8} + \dfrac{7}{12} + \dfrac{8}{16} + \dfrac{9}{20} + ... + \dfrac{76}{288} \bigg) - \Big(1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \frac{1}{5} + ... + \dfrac{1}{72} \bigg) \bigg] \cdot 2} =$

$= \sqrt{ \bigg(\dfrac{5}{4} - \dfrac{ ^{4)} 1}{1} + \dfrac{6}{8} - \dfrac{^{4)}1}{2} + \dfrac{7}{12} - \dfrac{^{4)}1}{3} + \dfrac{8}{16} - \dfrac{^{4)}1}{4} + \dfrac{9}{20} - \dfrac{^{4)} 1}{5} + ... + \dfrac{76}{288} - \dfrac{^{4)}1}{72}\bigg) \cdot 2}$

$= \sqrt{\bigg( \dfrac{5 - 4}{4} + \dfrac{6 - 4}{2 \cdot 4} + \dfrac{7 - 4}{3 \cdot 4} + \dfrac{8 - 4}{4 \cdot 4} + \dfrac{9 - 4}{4 \cdot 5} + ... + \dfrac{76 - 4}{72 \cdot 4}\bigg) \cdot 2}$

$= \sqrt{\bigg( \dfrac{1}{4} + \dfrac{ \not2}{ \not2 \cdot 4} + \dfrac{\not3}{\not3 \cdot 4} + \dfrac{\not4}{\not4 \cdot 4} + ... + \dfrac{72}{72 \cdot 4}\bigg) \cdot 2}$

$= \sqrt{\bigg(\underbrace{\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4} + ... + \dfrac{1}{4}}_{72}\bigg) \cdot 2} = \sqrt{ \dfrac{72}{4}\cdot 2 }$

$= \sqrt{36} = \bf 6$