Question

иррациональность .................​

Answer 1

$\displaystyle\bf\\1)\\\\\frac{1}{\sqrt[4]{8} } =\frac{1\cdot\sqrt[4]{2} }{\sqrt[4]{8}\cdot\sqrt[4]{2} } =\frac{\sqrt[4]{2} }{\sqrt[4]{16} } =\frac{\sqrt[4]{2} }{\sqrt[4]{2^{4} } } =\frac{\sqrt[4]{2} }{2} \\\\\\2)\\\\\frac{6}{\sqrt[3]{5} +1} =\frac{6\cdot(\sqrt[3]{25}-\sqrt[3]{5} +1) }{(\sqrt[3]{5} +1)(\sqrt[3]{25}-\sqrt[3]{5} +1) } =\frac{6\cdot(\sqrt[3]{25}-\sqrt[3]{5} +1) }{(\sqrt[3]{5} )^{3} +1^{3} } =$

$\displaystyle\bf\\=\frac{6\cdot(\sqrt[3]{25}-\sqrt[3]{5} +1) }{5+1}=\frac{6\cdot(\sqrt[3]{25}-\sqrt[3]{5} +1) }{6}=\sqrt[3]{25}-\sqrt[3]{5} +1$