Question

Выполните действие 1.2

Answer 1

$\sqrt[4]{6+\sqrt{20} } *\sqrt[4]{6-\sqrt{20} }=\sqrt[4]{(6+\sqrt{20} )(6-\sqrt{20} )}=\\=\sqrt[4]{6^2-(\sqrt{20} )^2}=\sqrt[4]{36-20}=\sqrt[4]{16}=2$

$(\sqrt{14}-3\sqrt{2})^2+6\sqrt{28}=(\sqrt{14})^2-6\sqrt{14}\sqrt{2}+(3\sqrt{2})^2+6\sqrt{28}=\\14-6\sqrt{28}+18+6\sqrt{28}=32$

$\sqrt[4]{4+\sqrt{15}}*\sqrt[4]{4-\sqrt{15}}=\sqrt[4]{(4+\sqrt{15})(4-\sqrt{15})}=\\=\sqrt[4]{4^2-(\sqrt{15})^2}=\sqrt[4]{16-15}=\sqrt[4]{1}=1$

$(3\sqrt{5}+\sqrt{15})^2-10\sqrt{27}=(3\sqrt{5})^2+6\sqrt{5}\sqrt{15}+(\sqrt{15})^2-10\sqrt{27}=\\=45+6\sqrt{75}+15-10\sqrt{27}=60+6\sqrt{25*3}-10\sqrt{9*3}=\\=60+30\sqrt{3}-30\sqrt{3}=60$