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@ Escribe cada radical en la forma más simple posil \( \begin{array}{ll}\text { a) } \sqrt[3]{a^{6}} & \begin{array}{ll}\text { d) } \sqrt{16 v^{6} w^{9}}= & \text { b) } \sqrt[3]{m^{4} n^{6}}= \\ \text { g) } \sqrt{\frac{125 x^{3} y^{6}}{49 x y}}= & \text { e) } \sqrt[3]{625 x^{2} y^{3}}= \\ \text { i) } \sqrt[5]{\frac{64 x^{6} y^{3}}{5 m^{3}}}= & \text { h) } \sqrt[3]{\frac{100 a^{3} b^{6}}{16 a b}}= \\ & \text { k) } \sqrt[3]{\frac{24 x^{3} y^{6}}{729}}=\end{array}\end{array}> \)

Ask by Luna Gordon. in Colombia
Mar 11,2025

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a) \( \sqrt[3]{a^{6}} = a^{2} \) b) \( \sqrt[3]{m^{4} n^{6}} = m\sqrt[3]{m} \cdot n^{2} \) d) \( \sqrt{16 v^{6} w^{9}} = 4v^{3}w^{4}\sqrt{w} \) e) \( \sqrt[3]{625 x^{2} y^{3}} = 5\sqrt[3]{5x^{2}} \cdot y \) g) \( \sqrt{\frac{125 x^{3} y^{6}}{49 x y}} = \frac{5x^{2}y^{3}\sqrt{5y}}{7} \) h) \( \sqrt[3]{\frac{100 a^{3} b^{6}}{16 a b}} = \frac{ab^{2}\sqrt[3]{400ab}}{4} \) i) \( \sqrt[5]{\frac{64 x^{6} y^{3}}{5 m^{3}}} = \frac{2x\sqrt[5]{1250xy^{3}m^{3}}}{5} \) k) \( \sqrt[3]{\frac{24 x^{3} y^{6}}{729}} = \frac{2\sqrt[3]{81} \cdot xy^{2}}{27} \)

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Bonus Knowledge

Para simplicidad, vamos a simplificar cada radical por separado. a) \( \sqrt[3]{a^{6}} = a^{2} \) (ya que \( a^{6} = (a^{2})^{3} \)) b) \( \sqrt[3]{m^{4} n^{6}} = mn^{2} \sqrt[3]{m} \) (una parte es un cubo perfecto) d) \( \sqrt{16 v^{6} w^{9}} = 4v^{3}w^{4} \) (cada exponente se divide por 2) e) \( \sqrt[3]{625 x^{2} y^{3}} = 5y \sqrt[3]{25x^{2}} \) (sacamos la raíz del cubo) g) \( \sqrt{\frac{125 x^{3} y^{6}}{49 x y}} = \frac{5y^3}{7} \sqrt{x^2} = \frac{5y^3}{7} x \) h) \( \sqrt[3]{\frac{100 a^{3} b^{6}}{16 a b}} = \frac{5b^2}{2} a^{\frac{2}{3}} \) i) \( \sqrt[5]{\frac{64 x^{6} y^{3}}{5 m^{3}}} = \frac{2x^{\frac{6}{5}}y^{\frac{3}{5}}}{5^{\frac{1}{5}}m^{\frac{3}{5}}} \) k) \( \sqrt[3]{\frac{24 x^{3} y^{6}}{729}} = \frac{2}{9}xy^{2} \sqrt[3]{24} \) (sacamos el cubo perfecto) Espero que esto te haya ayudado a simplificar esos radicales de la forma más sencilla posible. Si tienes más preguntas, ¡no dudes en preguntar!

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