Question

Aratati ca nr. n = 4 supra 3 - 4 supra 3 inmultit [4 supra 3 -(0,(3) la puterea a 2 ]:4 8 supra 9

Answer 1

Salutare!

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\Big[\dfrac{4}{3}-(0,(3))^{2}\Big]:4\dfrac{8}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\Big[\dfrac{4}{3}-\Big(\dfrac{3}{9}\Big)^{2}\Big]:\dfrac{4\cdot9+8}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\Big[\dfrac{4}{3}-\Big(\dfrac{\not3}{\not9}\Big)^{2}\Big]:\dfrac{44}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\Big[\dfrac{4}{3}-\Big(\dfrac{1}{3}\Big)^{2}\Big]:\dfrac{44}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\Big(\dfrac{4}{3}-\dfrac{1}{9}\Big):\dfrac{44}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\Big(\dfrac{4\cdot3}{3\cdot3}-\dfrac{1}{9}\Big):\dfrac{44}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\dfrac{12-1}{9}:\dfrac{44}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\dfrac{11}{9}:\dfrac{44}{9}$

$\bf n = \dfrac{4}{3}- \dfrac{4}{3}\cdot\dfrac{11}{9}\cdot\dfrac{9}{44}$

$\bf n = \dfrac{4}{3}- \dfrac{\not4}{3}\cdot\dfrac{\not11}{\not9}\cdot\dfrac{\not9}{\not44}$

$\bf n = \dfrac{4}{3}- \dfrac{1}{3}\cdot\dfrac{1}{1}\cdot\dfrac{1}{1}$

$\bf n = \dfrac{4}{3}- \dfrac{1}{3}$

$\bf n = \dfrac{4-1}{3}$

$\bf n = \dfrac{3}{3}$

$\bf n = \dfrac{\not3}{\not3}$

$\boxed{\bf n = 1}$