Question

dau coroana va rog ajutor

Answer 1

a)

$a = \bigg(\dfrac{1}{\sqrt{3}} + \dfrac{2}{\sqrt{12}} + \dfrac{3}{\sqrt{27}} + \dfrac{4}{\sqrt{48}}\bigg) : \dfrac{2}{\sqrt{3}} =$

$= \bigg(\dfrac{1}{\sqrt{3}} + \dfrac{\not2}{\not2\sqrt{3}} + \dfrac{\not3}{\not3\sqrt{3}} + \dfrac{\not4}{\not4\sqrt{3}}\bigg) \cdot \dfrac{\sqrt{3}}{2}$

$= \bigg(\dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{3}} + \dfrac{1}{\sqrt{3}}\bigg) \cdot \dfrac{\sqrt{3}}{2} =$

$= \dfrac{4}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{4}{2} = \bf 2$

b)

$b = \dfrac{^{12)} 1}{3} + \dfrac{^{6)}1}{6} + \dfrac{^{4)}1}{9} + \dfrac{^{3)}1}{12} + \dfrac{^{2)}1}{18} = \dfrac{12 + 6 + 4 + 3 + 2}{36} =$

$= \dfrac{27^{(9} }{36} = \bf \dfrac{3}{4}$

$N = \bigg(\dfrac{1}{a} - 2 \cdot b\bigg)^{2022} = \bigg(\dfrac{1}{2} - 2 \cdot \dfrac{3}{4}\bigg)^{2022} = \bigg(\dfrac{1}{2} - \dfrac{3}{2}\bigg)^{2022}$

$= \bigg(\dfrac{1-3}{2}\bigg)^{2022} = \bigg(-\dfrac{2}{2}\bigg)^{2022} = \bigg(-1\bigg)^{2022} = 1^{2022} = \bf 1$

$\implies N = 1$