Question

Что с первой дробью делать ума не приложу. $\frac{25\sqrt[4]{2} + 2\sqrt{5}}{\sqrt{250} + 5\sqrt[4]{8}} - \sqrt{\frac{\sqrt{2}}{5} + \frac{5}{\sqrt{2}} + 2} = -1 \\ \\\frac{25\sqrt[4]{2} + 2\sqrt{5}}{\sqrt{250} + 5\sqrt[4]{8}} - \sqrt{\frac{2 + 25 + 2 \cdot 5\sqrt{2}}{5\sqrt{2}}} = -1 \\ \\\frac{25\sqrt[4]{2} + 2\sqrt{5}}{\sqrt{250} + 5\sqrt[4]{8}} - \sqrt{\frac{27 + 10\sqrt{2}}{5\sqrt{2}}} = -1 \\\\\frac{25\sqrt[4]{2} + 2\sqrt{5}}{\sqrt{250} + 5\sqrt[4]{8}} - \sqrt{\frac{(5 + \sqrt{2})^{2}}{5\sqrt{2}}} = -1$

Answer 1

$\sqrt{\frac{\sqrt{2} }{5} +\frac{5}{\sqrt{2} }+2 } =\sqrt{(\frac{\sqrt[4]{2} }{\sqrt{5} }+\frac{\sqrt{5} }{\sqrt[4]{2} } )^2} =|\frac{\sqrt[4]{2} }{\sqrt{5} }+\frac{\sqrt{5} }{\sqrt[4]{2} } |=\frac{\sqrt[4]{2} }{\sqrt{5} }+\frac{\sqrt{5} }{\sqrt[4]{2} }$

Тогда

$\frac{25\sqrt[4]{2}+2\sqrt{5} }{\sqrt{250}+5\sqrt[4]{8} } =\frac{5^2\cdot 2^{\frac{1}{4}}+2\cdot 5^{\frac{1}{2}}}{5\cdot 5^{\frac{1}{2}}\cdot 2^{\frac{1}{2}}+5\cdot 2^{\frac{3}{4}}} =\frac{5^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}\cdot (5^{\frac{3}{2}}+2^{\frac{3}{4}})}{5\cdot 2^{\frac{1}{2}}(5^{\frac{1}{2}}+2^{\frac{1}{4}}) } =$

$=\frac{(5^{\frac{1}{2}})^3+(2^{\frac{1}{4}})^3}{5^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}(5^{\frac{1}{2}}+2^{\frac{1}{4}}) } =\frac{5-5^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}+2^{\frac{1}{2}}}{5^{\frac{1}{2}}\cdot 2^{\frac{1}{4}}} =\frac{\sqrt{5} }{\sqrt[4]{2} } -1+\frac{\sqrt[4]{2} }{\sqrt{5} }$

Итак,

$\frac{25\sqrt[4]{2}+2\sqrt{5} }{\sqrt{250}+5\sqrt[4]{8} } -\sqrt{\frac{\sqrt{2} }{5} +\frac{5}{\sqrt{2} }+2 } =\frac{\sqrt{5} }{\sqrt[4]{2} } -1+\frac{\sqrt[4]{2} }{\sqrt{5} }-(\frac{\sqrt[4]{2} }{\sqrt{5} }+\frac{\sqrt{5} }{\sqrt[4]{2} })=-1$

что и требовалось доказать