ГДЗ по алгебре 9 класс Макарычев Задание 196

\[\boxed{\text{196\ (196).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ x > 0\]

\[x^{6} = x^{3 \cdot 2} = \left( x^{3} \right)^{2},\ \ \]

\[x^{5} = x^{2,5 \cdot 2} = \left( x^{2,5} \right)^{2};\]

\[x^{- 8} = x^{- 4 \cdot 2} = \left( x^{- 4} \right)^{2},\ \ \]

\[x^{- 1} = x^{- 0,5 \cdot 2} = \left( x^{- 0,5} \right)^{2};\]

\[x = x^{0,5 \cdot 2} = \left( x^{0,5} \right)^{2},\ \ \]

\[x^{\frac{1}{3}} = x^{\frac{1}{6} \cdot 2} = \left( x^{\frac{1}{6}} \right)^{2}.\]

\[\textbf{б)}\ \ y^{6} = y^{2 \cdot 3} = \left( y^{2} \right)^{3},\ \ \]

\[y^{7} = y^{\frac{7}{3} \cdot 3} = \left( y^{\frac{7}{3}} \right)^{3};\]

\[y = y^{\frac{1}{3} \cdot 3} = \left( y^{\frac{1}{3}} \right)^{3},\ \ \]

\[y^{\frac{1}{2}} = y^{\frac{1}{6} \cdot 3} = \left( y^{\frac{1}{6}} \right)^{3};\]

\[y^{- 1,5} = y^{- 0,5 \cdot 3} = \left( y^{- 0,5} \right)^{3},\ \ \]

\[y^{0,2} = y^{\frac{1}{5}} = \left( y^{\frac{1}{15}} \right)^{3};\]

\[y^{- \frac{2}{9}} = y^{- \frac{2}{27} \cdot 3} = \left( y^{- \frac{2}{27}} \right)^{3}\]