Question

• Alguém que seja bom em matemática pode por gentileza me ajudar?:')​

Answer 1

Resposta:

Explicação passo-a-passo:

a) $\left( \frac{a^2b}{c} \right )^3\cdot\left( \frac{c^}{a^3} \right )^2\cdot\left( \frac{1}{b} \right )^{-2}=\frac{a^6b^3}{c^3} \cdot\frac{c^2}{a^6} \cdot \frac{b^2}{1} = \frac{a^6b^5c^2}{a^6c^3}=\frac{b^5}{c}$

b) $\left( \dfrac{xy^2}{2} \right)^4\left( \dfrac{x^2y}{4} \right)^{-2}=\dfrac{x^4y^8}{16}\cdot\dfrac{16}{x^4y^2} =y^6$

c)

$\left( \dfrac{3xy}{4} \right)^{-3}\left( \dfrac{2x^2y^2}{3} \right)^{2}:\left(\dfrac{16x}{9y}\right)=\dfrac{64}{27x^3y^3}\cdot\dfrac{4x^4y^4}{9}:\dfrac{16x}{9y}=\dfrac{256x^4y^4}{243x^3y^3}:\dfrac{16x}{9y}=\\\\\\\dfrac{256xy}{243}:\dfrac{16x}{9y}=\dfrac{2304xy^2}{3888x}=\dfrac{16y^2}{27}$