Решебник по алгебре 9 класс Макарычев Задание 193

\[\boxed{\text{193\ (193).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ c^{\frac{1}{2}} \cdot c^{\frac{1}{3}} = c^{\frac{1}{2} + \frac{1}{3}} = c^{\frac{5}{6}};\]

\[\textbf{б)}\ b^{- \frac{1}{3}} \cdot b^{\frac{1}{2}} = b^{\frac{1}{2} - \frac{1}{3}} = b^{\frac{1}{6}};\]

\[\textbf{в)}\ a^{\frac{2}{3}} \cdot a^{\frac{1}{6}} = a^{\frac{2}{3} + \frac{1}{6}} = a^{\frac{5}{6}};\]

\[\textbf{г)}\ d^{5} \cdot d^{\frac{1}{2}} = d^{5 + 0,5} = d^{5.5};\]

\[\textbf{д)}\ x^{\frac{1}{2}}\ :x^{\frac{3}{2}} = x^{\frac{1}{2} - \frac{3}{2}} = x^{- 1};\]

\[\textbf{е)}\ y^{\frac{5}{6}}\ :y^{\frac{1}{3}} = y^{\frac{5}{6} - \frac{1}{3}} = y^{\frac{1}{2}};\]

\[\textbf{ж)}\ z^{\frac{1}{5}}\ :z^{- \frac{1}{2}} = z^{\frac{1}{5} + \frac{1}{2}} = z^{\frac{7}{10}};\]

\[\textbf{з)}\ m^{\frac{1}{3}}\ :m^{2} = m^{\frac{1}{3} - 2} = m^{- \frac{5}{3}};\]

\[\textbf{и)}\ \left( b^{\frac{1}{2}} \right)^{\frac{1}{3}} = b^{\frac{1}{6}};\]

\[к)\ \left( a^{\frac{3}{2}} \right)^{\frac{4}{9}} = a^{\frac{3}{2} \cdot \frac{4}{9}} = a^{\frac{2}{3}};\]

\[л)\ \left( c^{- \frac{1}{2}} \right)^{\frac{1}{3}} = c^{- \frac{1}{6}};\]

\[м)\ \left( p^{3} \right)^{- \frac{2}{9}} = p^{- \frac{2}{3}}\text{.\ }\]

\[\boxed{\mathbf{193}\text{.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[y = ax²\]

\[\textbf{а)}\ ax^{2} \geq 0\ \ \ при\ \ a \geq 0;\]

\[\textbf{б)}\ \ ax^{2} \leq 0\ \ при\ \ a \leq 0\text{.\ }\]