Question

1. $\sqrt[3]{3-\sqrt{10}} \cdot \sqrt[6]{19+6\sqrt{10}}$

2. $\frac{1}{y} - \frac{1}{x}$, если $\frac{\sqrt{2}y - \sqrt{2}x}{xy}=\sqrt{32}$

3. $\frac{4\sqrt[4]{x} + x\sqrt{2}}{2\sqrt[4]{x} + \sqrt{2x}} + \sqrt{4+x-4\sqrt{x}}$ . В ответ напишите при х=$\frac{81}{64}$

Answer 1

$1)\sqrt[3]{3-\sqrt{10}}*\sqrt[6]{19+6\sqrt{10}}=\sqrt[3]{3-\sqrt{10}}*\sqrt[6]{9+2*3*\sqrt{10}+10}=\sqrt[3]{3-\sqrt{10}}*\sqrt[6]{(3+\sqrt{10})^{2}}=\sqrt[3]{3-\sqrt{10}}*\sqrt[3]{3+\sqrt{10}} =\sqrt[3]{(3-\sqrt{10})(3+\sqrt{10})}=\sqrt[3]{3^{2}-(\sqrt{10})^{2}}=\sqrt[3]{9-10}=\sqrt[3]{-1}=-1$

$2)\frac{\sqrt{2}y-\sqrt{2}x}{xy}=\sqrt{32}\\\\\frac{\sqrt{2}(y-x) }{xy}=\sqrt{32}\\\\\frac{y-x}{xy}=\frac{\sqrt{32}}{\sqrt{2}}\\\\\frac{y-x}{xy}=4\\\\\\\frac{1}{y}-\frac{1}{x}=\frac{x-y}{xy}=-\frac{y-x}{xy}=-4$

$3)\frac{4\sqrt[4]{x}+x\sqrt{2}}{2\sqrt[4]{x}+\sqrt{2x}} +\sqrt{4+x-4\sqrt{x}}=\frac{\sqrt{2}*\sqrt[4]{x}*(2\sqrt{2}+\sqrt[4]{x^{3}})}{\sqrt{2}*\sqrt[4]{x}*(\sqrt{2}+\sqrt[4]{x})}+\sqrt{4-4\sqrt{x} +x}=\frac{(\sqrt{2})^{3}+(\sqrt[4]{x})^{3}}{\sqrt{2}+\sqrt[4]{x}}+\sqrt{(2-\sqrt{x})^{2}}=\frac{(\sqrt{2}+\sqrt[4]{x})(2-\sqrt{2}*\sqrt[4]{x}+\sqrt[4]{x^{2}})}{\sqrt{2}+\sqrt[4]{x}}+2-\sqrt{x}=2-4\sqrt[4]{x}+\sqrt{x}+2-\sqrt{x}=4-\sqrt[4]{4x}$

$x=\frac{81}{64} \Rightarrow 4-\sqrt[4]{4*\frac{81}{64}}=4-\sqrt[4]{\frac{81}{16}}=4-\sqrt[4]{(\frac{3}{2})^{4}}=4-1,5=2,5$