Решебник по алгебре 9 класс Макарычев Задание 161

 

\[\boxed{\text{161\ (161).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \sqrt[9]{512} = \sqrt[9]{2^{9}} = 2\]

\[\textbf{б)}\ \sqrt[3]{1331} = \sqrt[3]{11^{3}} = 11\]

\[\textbf{в)}\ \sqrt[8]{0} = 0\]

\[\textbf{г)}\ \sqrt[5]{- 243} = \sqrt[5]{- 3^{5}} = - 3\]

\[\textbf{д)}\ \sqrt[4]{\frac{16}{625}} = \sqrt[4]{\frac{2^{4}}{5^{4}}} = \frac{2}{5}\]

\[\textbf{е)}\ \sqrt[6]{\frac{64}{729}} = \sqrt[6]{\frac{2^{6}}{3^{6}}} = \frac{2}{3}\]

\[\boxed{\text{161.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \ c = 5;b = 0;\]

\[A(5;0):\]

\[0 = 25a + 5\]

\[25a = - 5\]

\[a = - \frac{1}{5}.\]

\[a = - 0,2;\ \ b = 0;\ \ c = 5.\]

\[\textbf{б)}\ c = - 3;B(2; - 5)\]

\[- \frac{b}{2a} = 2;\ \ b = - 4a;\]

\[y = ax^{2} - 4ax - 3\]

\[- 5 = 4a - 8a - 3\]

\[- 4a = - 2\]

\[a = 0,5.\]

\[b = - 4a = - 4 \cdot 0,5 = - 2.\]

\[a = 0,5;\ \ b = - 2;\ \ c = - 3.\]

\[\textbf{в)}\ A( - 3;0);\ \ B( - 2;2);\ \]

\[\text{\ C}( - 4;2):\]

\[\left\{ \begin{matrix} 0 = 9a - 3b + c\ \ \\ 2 = 4a - 2b + c\ \ \\ 2 = 16a - 4b + c \\ \end{matrix} \right.\ \]

\[12a - 10a - 4 = 0\]

\[2a = 4\]

\[a = 2.\]

\[b = 5 \cdot 2 + 2 = 12.\]

\[0 = 9 \cdot 2 - 3 \cdot 12 + c\]

\[c = 18.\]

\[a = 2;\ \ b = 12;\ \ c = 18.\]

\[\textbf{г)}\ c = 3;\ \ \ A( - 2;5)\]

\[- \frac{b}{2a} = - 2;\ \ \ b = 4a;\]

\[y = ax^{2} + 4ax + 3\]

\[5 = 4a - 8a + 3\]

\[- 4a = 2\]

\[a = - 0,5.\]

\[b = 4a = 4 \cdot ( - 0,5) = - 2.\]

\[a = - 0,5;\ \ b = - 2;\ \ c = 3.\]