Question

CPCAR - O inverso de raiz de x/y vezes raiz cúbica de x/y, com x > 0 e y > 0, é igual a:

Answer 1

Resposta:

Explicação passo-a-passo:

(\frac{x}{y} .(\frac{x}{y} )^{\frac{1}{3} } )^{\frac{1}{2} } =\\\\

(\frac{x}{y})^{\frac{1}{2} }.(\frac{x}{y} )^{\frac{1}{6} }=\\

(\frac{x}{y})^{\frac{1}{2} +\frac{1}{6} } =\\

(\frac{x}{y})^{\frac{2}{3}}=\\\\

\frac{\sqrt[3]{x^{2} } }{\sqrt[3]{y^{2} } } =\\\\\\\\\frac{\sqrt[3]{x^{2} } }{y\frac{2}{3} } =\\\\\

\frac{\sqrt[3]{x^{2} } }{y^{1} .y^{-\frac{1}{3} } }=\\\\\\\frac{\sqrt[3]{x^{2} }.y^{\frac{1}{3} }  }{y} =\\

\\\frac{\sqrt[3]{x^{2}} .\sqrt[3]{y} }{y}  =\\\\\frac{\sqrt[3]{x^{2} y } }{y}

Answer 2

$(\frac{x}{y} .(\frac{x}{y} )^{\frac{1}{3} } )^{\frac{1}{2} } =\\\\</p><p></p><p>(\frac{x}{y})^{\frac{1}{2} }.(\frac{x}{y} )^{\frac{1}{6} }=\\</p><p></p><p>(\frac{x}{y})^{\frac{1}{2} +\frac{1}{6} } =\\</p><p></p><p>(\frac{x}{y})^{\frac{2}{3}}=\\\\</p><p></p><p>\frac{\sqrt[3]{x^{2} } }{\sqrt[3]{y^{2} } } =\\\\\\\\\frac{\sqrt[3]{x^{2} } }{y\frac{2}{3} } =\\\\\</p><p></p><p>\frac{\sqrt[3]{x^{2} } }{y^{1} .y^{-\frac{1}{3} } }=\\\\\\\frac{\sqrt[3]{x^{2} }.y^{\frac{1}{3} }  }{y} =\\</p><p></p><p>\\\frac{\sqrt[3]{x^{2}} .\sqrt[3]{y} }{y}  =\\\\\frac{\sqrt[3]{x^{2} y } }{y}</p><p></p><p></p><p>$