Question

Решение квадратных корней

Answer 1

$1)\; \; \frac{3}{\sqrt[3]5}=\frac{3\cdot \sqrt[3]{5^2}}{\sqrt[3]{5}\cdot \sqrt[3]{5^2}}=\frac{3\sqrt[3]{25}}{\sqrt[3]{5^3}}=\frac{3\, \sqrt[3]{25}}{5}\\\\\\2)\; \; \frac{6}{\sqrt[3]5+1}=\frac{6\cdot (\sqrt[3]{5^2}-\sqrt[3]5+1)}{(\sqrt[3]{5}+1)(\sqrt[3]{5^2}-\sqrt[3]5+1)}=\frac{6\cdot  (\sqrt[3]{5^2}-\sqrt[3]5+1)}{(\sqrt[3]5)^3+1^2}=\frac{6\cdot  (\sqrt[3]{5^2}-\sqrt[3]5+1)}{5+1}=\sqrt[3]{5^2}-\sqrt[3]5+1$

$3)\; \; \frac{3}{\sqrt[3]{16}+\sqrt[3]4+1}=\frac{3\cdot (\sqrt[3]{4}-1)}{(\sqrt[3]{4^2}+\sqrt[3]4+1) (\sqrt[3]{4}-1)}=\frac{3\cdot (\sqrt[3]{4}-1)}{(\sqrt[3]4)^3-1^3}=\frac{3\cdot (\sqrt[3]{4}-1)}{4-1}=\sqrt[3]{4}-1$