Question

Помогите пжл, очень срочно,буду очень благодарен

Answer 1

$\frac{\sqrt[4]{x*\sqrt[3]x}}{\sqrt[3]x}=\frac{(x^{\frac{4}{3}})^{\frac{1}{4}}}{x^{\frac{1}{3}}}=1\\\\\frac{\sqrt[5]{y^4}-9}{\sqrt[5]{y^2}-3}-\sqrt[5]{y^2}=\frac{(y^{\frac{2}{5}}-3)(y^{\frac{2}{5}}+3)}{y^{\frac{2}{5}}-3}-\sqrt[5]{y^2}=y^{\frac{2}{5}}+3-y^{\frac{2}{5}}=3\\\\\frac{10}{\sqrt[5]8}=\frac{10*8^{\frac{4}{5}}}{8^{\frac{1}{5}}*8^{\frac{4}{5}}}=\frac{10*8^{\frac{4}{5}}}{8}=1,25*8^{\frac{4}{5}}=5*\sqrt[5]4$

Answer 2

$\frac{ \sqrt[4]{x* \sqrt[3]{x} } }{ \sqrt[3]{x} } = \frac{(x*x^ \frac{1}{3})^ \frac{1}{4} }{x^ \frac{1}{3} }= \frac{x^ \frac{1}{4}*x^ \frac{1}{12} }{x^ \frac{1}{3} }= \frac{x^ \frac{1}{3} }{x^ \frac{1}{3} }=1 [tex] \frac{ \sqrt[5]{y^4}-9 }{ \sqrt[5]{x^2}-3}- \sqrt[5]{x^2}= \frac{( \sqrt[5]{x^2}-3)( \sqrt[5]{x^2}+3) }{ \sqrt[5]{x^2}-3}- \sqrt[5]{x^2}= \sqrt[5]{x^2}+3- \sqrt[5]{x^2}=3$
$\frac{10}{ \sqrt[5]{8}}= \frac{10* (\sqrt[5]{8})^4}{8}=1.25* 8^ \frac{4}{5}$