Question

Избавьтесь от иррациональности. Задание 1

Answer 1

$frac{3}{sqrt[4]{27}}=frac{3sqrt[4]{27^3}}{sqrt[4]{27}cdot sqrt[4]{27^3}}=frac{3sqrt[4]{3^9}}{27}=frac{27sqrt[4]{3}}{27}=sqrt[4]{3}\\frac{5}{sqrt[3]{5x}}=frac{5sqrt[3]{(5x)^2}}{5x}=frac{sqrt[3]{25x^2}}{x}\\frac{7}{sqrt[4]{49x}}=frac{7sqrt[4]{(49x)^3}}{49x}=frac{sqrt[4]{7^6x^3}}{7x}=frac{7sqrt[4]{7^2x^3}}{7x}=frac{sqrt[4]{49x^3}}{x}$

Answer 2

1)$frac{3}{sqrt[4]{27}} =frac{3*27^{3/4}}{sqrt[4]{27}*27^{3/4}}} = frac{3*3^{3^{3/4}}}{27}} = frac{3*3^{9/4}}{27}} = frac{3^3*3^{1/4}}{3^3}} =sqrt[4]{3}$

2)$frac{5}{sqrt[3]{5x}} =frac{5*(5x)^{2/3}}{sqrt[3]{5x}*(5x)^{2/3}}=frac{5*(25x^2)^{1/3}}{5x}=frac{sqrt[3]{25x^2}}{x}$

3)$frac{7}{sqrt[4]{49x}} =frac{7}{sqrt{7}sqrt[4]{x}} =frac{7sqrt{7}*x^{3/4}}{sqrt{7}sqrt{7}sqrt[4]{x}*x^{3/4}} =frac{sqrt{7}*x^{3/4}}{x} = frac{sqrt[4]{49x^3}}{x}$