Question

2, Inversul numărului 2 (5-3√3)² √(2√3-4) 2 2 este:
va rog toate 3
DAU COROANAA VA TOGG​

Answer 1

Răspuns:

$C. \ \sqrt3-1$

Explicație pas cu pas:

$\sqrt{(5-3\sqrt3)^2}=|5-3\sqrt{3}|=3\sqrt3-5\\3\sqrt3=\sqrt{3^2\cdot3}=\sqrt{27} > 5\\\sqrt{(2\sqrt3-4)^2}=|2\sqrt3-4|=4-2\sqrt3\\2\sqrt3=\sqrt{2^2\cdot3}=\sqrt{12} < 4$

$a=\frac{1}{3\sqrt3-5} -\frac{2}{4-2\sqrt3}=\frac{3\sqrt3+5}{2} -\frac{2(4+2\sqrt3)}{4} =\frac{6\sqrt3+10-8-4\sqrt3}{4} =\frac{2\sqrt3+2}{4} =\frac{\sqrt3+1}{2}$

$\frac{1}{a} =\frac{1}{\frac{\sqrt3+1}{2} } =\frac{2}{\sqrt3 +1} =\frac{2(\sqrt3-1)}{2} =\sqrt3-1$

Answer 2

Explicație pas cu pas:

3.

în fiecare paranteză sunt 72 de termeni, pe care îi grupăm câte 2 (câte unul din fiecare paranteză:

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

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

$= \sqrt{\Big(\underbrace{\dfrac{1}{4} + \dfrac{1}{4} + ... + \dfrac{1}{4}}_{72} \Big) \cdot 2} = \sqrt{\dfrac{72}{4} \cdot 2} = \sqrt{36} = \bf 6$