Question

Избавьтесь от иррациональности. Задание 1

Answer 1

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

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