Question

Помогите пожалуйста решить алгебру (9 класс)​

Answer 1

$</p><p>\frac{ \sqrt[6]{a \sqrt[3]{ {a}^{ - 1} } } }{ {a}^{ - \frac{2}{9} } } = \frac{ \sqrt[6]{a \times a {}^{ - \frac{1}{3} } } }{{a}^{ - \frac{2}{9} }} = \frac{ \sqrt[6]{ {a}^{ \frac{2}{3} } } }{{a}^{ - \frac{2}{9} }} = \frac{(a {}^{ \frac{2}{3} } ) {}^{ \frac{1}{6} } }{{a}^{ - \frac{2}{9} }} = \frac{a {}^{ \frac{1}{9} } }{{a}^{ - \frac{2}{9} }} = a {}^{ \frac{1}{9} - ( - \frac{2}{9} ) } = a {}^{ \frac{3}{9} } = a {}^{ \frac{1}{3} } = \sqrt[3]{a} </p><p> \\ \\ \frac{ \sqrt[4]{ {x}^{3} \sqrt[3]{x} } }{x {}^{ \frac{1}{3} } } = \frac{ \sqrt[4]{x {}^{3} \times x {}^{ \frac{1}{3} } } }{x {}^{ \frac{1}{3} } } = \frac{ \sqrt[4]{x {}^{ \frac{10}{3} } } }{x {}^{ \frac{1}{3} } } = \frac{( {x}^{ \frac{10}{3} }) {}^{ \frac{1}{4} } }{x {}^{ \frac{1}{3} } } = \frac{x {}^{ \frac{5}{6} } }{ x {}^{ \frac{1}{3} } } = x {}^{ \frac{5}{6} - \frac{2}{6} } = x {}^{ \frac{3}{6} } = x {}^{ \frac{1}{2} } = \sqrt{x}$