Question

(EPCAR) O inverso de ✓((x/y) ³✓(y/x)) , com x>0 e y>0, é igual a:​

Answer 1

\sqrt[3]{y} } $\sqrt{ \frac{x}{y}  \sqrt[3]{ \frac{y}{x} } } = \sqrt{\sqrt[3]{  \frac{ x^{3} }{ y^{3} } .\frac{y}{x} } } = \sqrt[6]{  \frac{ x^{2} }{ y^{2} } }= \sqrt[3]{  \frac{x}{y} }=  \frac{ \sqrt[3]{x} }{ \sqrt[3]{y} } \\\\ O  inverso é:   \\\\\frac{ \sqrt[3]{y} }{ \sqrt[3]{x} } \frac{ \sqrt[3]{y} }{ \sqrt[3]{x} } =\frac{ \sqrt[3]{y} .\sqrt[3]{ x^{2} }}{ \sqrt[3]{x}.\sqrt[3]{ x^{2} } } =\frac{ \sqrt[3]{ x^{2}y }}{ x} }$

Answer 2

$\sqrt{ \frac{x}{y}~.~\sqrt[3]{\frac{x}{y}} }} = k$

Temos que o inverso disso será 1/k:

$\frac{1}{k} = \frac{1}{\sqrt{ \frac{x}{y}~.~\sqrt[3]{\frac{x}{y}}}} = \frac{\sqrt{ \frac{x}{y}~.~\sqrt[3]{\frac{x}{y}} }}{(\sqrt{ \frac{x}{y}~.~\sqrt[3]{\frac{x}{y}} })^2} = \frac{\sqrt{ \frac{x}{y}~.~\sqrt[3]{\frac{x}{y}} }}{\frac{x}{y}~.~\sqrt[3]{\frac{x}{y}} }} = \frac{\sqrt{ \frac{x}{y}~.~\sqrt[3]{\frac{x}{y}} }.(\sqrt[3]{\frac{x}{y}})^2}{\frac{x}{y}.\sqrt[3]{\frac{x}{y}}.(\sqrt[3]{\frac{x}{y}})^2}=$

$\frac{\sqrt{ \frac{x}{y}~.~\sqrt[3]{\frac{x}{y}} }.(\sqrt[3]{\frac{x}{y}})^2}{\frac{x}{y}.\frac{x}{y}} =\frac{\sqrt{ \frac{x}{y}}.{\sqrt{\sqrt[3]{\frac{x}{y}} }.(\sqrt[3]{\frac{x}{y}})^2}}{\frac{x^2}{y^2}} = \frac{\sqrt{ \frac{x}{y}}.{\sqrt[6]{\frac{x}{y} }.\sqrt[3]{\frac{x^2}{y^2}}}}{\frac{x^2}{y^2}} =$

Transformando todas as raízes em expoentes e realizando o jogo de expoentes:

${x^\frac{-4}{6}}.y^\frac{4}{6} = x^\frac{-2}{3} .y^\frac{2}{3} = \frac{\sqrt[3]{y^2} }{\sqrt[3]{x^2} } =  \frac{\sqrt[3]{y^2} }{\sqrt[3]{x^2} } . \frac{\sqrt[3]{x}}{\sqrt[3]{x}}  = \boxed{\boxed{\boxed{\frac{\sqrt[3]{y^2.x} }{x}}}}$