Question

Calculează:
Vă rog!!!!​

Answer 1

$\it \bigg(\dfrac{2}{\sqrt3}\bigg)^{-2}=\bigg(\dfrac{\sqrt3}{2}\bigg)^{2}=\dfrac{3}{4}\\ \\ \\ \dfrac{\sqrt8}{9}-\dfrac{^{\sqrt2)}1}{\ \sqrt2}=\dfrac{\sqrt{4\cdot2}}{9}-\dfrac{\sqrt2}{2}=\dfrac{^{2)}2\sqrt2}{9}-\dfrac{^{9)}\sqrt2}{2}=\dfrac{4\sqrt2-9\sqrt2}{18}=-\dfrac{5\sqrt2}{18}$

Exercițiul  devine:

$\it \dfrac{\not3\ ^1}{4}\cdot\bigg(-\dfrac{5\sqrt2}{\not18\ _6}\bigg)=-\dfrac{5\sqrt2}{24}$