Question

Salut, in poza (●'◡'●)

Answer 1

Explicație pas cu pas:

1.a)

$\bigg(-\dfrac{8x^{3}}{10y}\bigg)^{2} = \bigg(\dfrac{4x^{3}}{5y}\bigg)^{2} = \dfrac{ {4}^{2} {(x^{3})}^{2} }{{5}^{2}{y}^{2} } = \dfrac{16 x^{6}}{{5}^{2}{y}^{2} }$

b)

$\bigg(\dfrac{2x - 1}{x + 2}\bigg)^{2} = \dfrac{{(2x)}^{2} - 2 \cdot 2x + 1}{ {x}^{2} + 2 \cdot 2x + 4} = \dfrac{4{x}^{2} - 4x + 1}{ {x}^{2} + 4x + 4}$

2.

$\bigg[\bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{5} : \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{3}\bigg]^{2} \cdot \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{4} =$

$= \bigg[\bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{5 - 3}\bigg]^{2} \cdot \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{4} = \bigg[\bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{2}\bigg]^{2} \cdot \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{4}$

$= \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{2 + 2} \cdot \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{4} = \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{4} \cdot \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{4}$

$= \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{4 + 4} = \bf \bigg(\dfrac{1 - 2x}{x - 4}\bigg)^{8}$