Question

Помогите пожалуйста

Answer 1

$\mathtt{(x+y\sqrt{\frac{y}{x}})^{\frac{2}{3}}*(\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}})^{-\frac{2}{3}}}$

упростим первый множитель: 

$\mathtt{(x+y\sqrt{\frac{y}{x}})^{\frac{2}{3}}=(x+\frac{y\sqrt{y}}{\sqrt{x}})^{\frac{2}{3}}=(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}})^{\frac{2}{3}}}$

упростим второй множитель: 

$\mathtt{(\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}})^{-\frac{2}{3}}=(\frac{\sqrt{xy}+(\sqrt{x}-\sqrt{y})^2}{\sqrt{x}(\sqrt{x}-\sqrt{y})})^{-\frac{2}{3}}=(\frac{x-\sqrt{xy}}{x-\sqrt{xy}+y})^{\frac{2}{3}}}$

так, 

$\mathtt{(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}})^{\frac{2}{3}}*(\frac{x-\sqrt{xy}}{x-\sqrt{xy}+y})^{\frac{2}{3}}=[\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}}*\frac{(x-\sqrt{xy})(\sqrt{x}+\sqrt{y})}{(\sqrt{x}+\sqrt{y})(x-\sqrt{xy}+y)}]^{\frac{2}{3}}=}\\\\\mathtt{[\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}}*\frac{\sqrt{x}(x-y)}{(\sqrt{x})^3+(\sqrt{y})^3}]^{\frac{2}{3}}=(x-y)^{\frac{2}{3}}=\sqrt[3]{(x-y)^2}}$