\[\boxed{\text{191\ (191).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \sqrt{1,3} = {1,3}^{\frac{1}{2}};\]
\[\textbf{б)}\ \sqrt[3]{7^{- 1}} = 7^{- \frac{1}{3}};\]
\[\textbf{в)}\ \sqrt[4]{\frac{2}{3}} = \left( \frac{2}{3} \right)^{\frac{1}{4}};\]
\[\textbf{г)}\ \sqrt[5]{\left( \frac{3}{2} \right)^{- 2}} = \sqrt[5]{\left( \frac{2}{3} \right)^{2}} = \left( \frac{2}{3} \right)^{\frac{2}{5}};\]
\[\textbf{д)}\ \sqrt[7]{a^{4}} = a^{\frac{4}{7}};\]
\[\textbf{е)}\ \frac{1}{\sqrt[4]{x^{3}}} = \frac{1}{x^{\frac{3}{4}}\text{\ \ }} = x^{- \frac{3}{4}};\]
\[\textbf{ж)}\ \sqrt[3]{a^{2} - b^{2}} = \left( a^{2} - b^{2} \right)^{\frac{1}{3}};\]
\[\textbf{з)}\ \sqrt[5]{(x - y)^{2}\ } = (x - y)^{\frac{2}{5}}.\]
\[\boxed{\text{191.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[y = ax²\]
\[\textbf{а)}\ (5;\ - 7):\ \ \ \]
\[- 7 = a \cdot 5^{2}\]
\[- 7 = 25a\]
\[a = - \frac{7}{25}\]
\[Ответ:при\ a = - \frac{7}{25}.\]
\[\textbf{б)}\ \left( - \sqrt{3};9 \right):\ \ \]
\[9 = a \cdot \left( - \sqrt{3} \right)^{2}\]
\[9 = 3a\]
\[a = 3\]
\[Ответ:при\ a = 3.\]
\[\textbf{в)}\ \left( - \frac{1}{2};\ - \frac{1}{2} \right):\ \ \]
\[- \frac{1}{2} = a \cdot \left( - \frac{1}{2} \right)^{2}\]
\[- \frac{1}{2} = a \cdot \frac{1}{4}\]
\[a = - \frac{1}{2}\ :\frac{1}{4}\]
\[a = - 2\]
\[Ответ:при\ a = - 2.\]
\[\textbf{г)}\ (100;10):\]
\[10 = a \cdot 100^{2}\]
\[10 = 10\ 000a\]
\[a = \frac{10}{10000}\]
\[a = 0,001\]
\[Ответ:при\ a = 0,001.\]