ГДЗ по алгебре 9 класс Макарычев Задание 191

 

\[\boxed{\text{191\ (191).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \sqrt{1,3} = {1,3}^{\frac{1}{2}};\]

\[\textbf{б)}\ \sqrt[3]{7^{- 1}} = 7^{- \frac{1}{3}};\]

\[\textbf{в)}\ \sqrt[4]{\frac{2}{3}} = \left( \frac{2}{3} \right)^{\frac{1}{4}};\]

\[\textbf{г)}\ \sqrt[5]{\left( \frac{3}{2} \right)^{- 2}} = \sqrt[5]{\left( \frac{2}{3} \right)^{2}} = \left( \frac{2}{3} \right)^{\frac{2}{5}};\]

\[\textbf{д)}\ \sqrt[7]{a^{4}} = a^{\frac{4}{7}};\]

\[\textbf{е)}\ \frac{1}{\sqrt[4]{x^{3}}} = \frac{1}{x^{\frac{3}{4}}\text{\ \ }} = x^{- \frac{3}{4}};\]

\[\textbf{ж)}\ \sqrt[3]{a^{2} - b^{2}} = \left( a^{2} - b^{2} \right)^{\frac{1}{3}};\]

\[\textbf{з)}\ \sqrt[5]{(x - y)^{2}\ } = (x - y)^{\frac{2}{5}}.\]

\[\boxed{\text{191.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[y = ax²\]

\[\textbf{а)}\ (5;\ - 7):\ \ \ \]

\[- 7 = a \cdot 5^{2}\]

\[- 7 = 25a\]

\[a = - \frac{7}{25}\]

\[Ответ:при\ a = - \frac{7}{25}.\]

\[\textbf{б)}\ \left( - \sqrt{3};9 \right):\ \ \]

\[9 = a \cdot \left( - \sqrt{3} \right)^{2}\]

\[9 = 3a\]

\[a = 3\]

\[Ответ:при\ a = 3.\]

\[\textbf{в)}\ \left( - \frac{1}{2};\ - \frac{1}{2} \right):\ \ \]

\[- \frac{1}{2} = a \cdot \left( - \frac{1}{2} \right)^{2}\]

\[- \frac{1}{2} = a \cdot \frac{1}{4}\]

\[a = - \frac{1}{2}\ :\frac{1}{4}\]

\[a = - 2\]

\[Ответ:при\ a = - 2.\]

\[\textbf{г)}\ (100;10):\]

\[10 = a \cdot 100^{2}\]

\[10 = 10\ 000a\]

\[a = \frac{10}{10000}\]

\[a = 0,001\]

\[Ответ:при\ a = 0,001.\]