ГДЗ по алгебре 9 класс Макарычев Задание 533

 

\[\boxed{\text{533\ (533).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} \frac{1}{x} + \frac{1}{y} = \frac{1}{6}\text{\ \ } \\ 2x - y = 5 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 6y + 6x = xy \\ 2x - y = 5\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x² - 23x + 30 = 0 \\ \end{matrix} \right.\ \]

\[D = 23^{2} - 4 \cdot 2 \cdot 30 = 289\]

\[x_{1,2} = \frac{23 \pm 17}{4} = 10;\frac{3}{2};\ \]

\[1)\ x_{1} = 10,\ \ y_{1} = 15;\]

\[2)\ x_{2} = 1,5;\ \ \ \ y_{2} = - 2.\]

\[\textbf{б)}\ \left\{ \begin{matrix} \frac{1}{x} - \frac{1}{y} = \frac{1}{20} \\ x + 2y = 14 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 20y - 20x = xy \\ x = 14 - 2y\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y² + 23y - 140 = 0 \\ x = 14 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[D = 23^{2} + 4 \cdot 140 = 1089 = 33^{2}\]

\[y_{1,2} = \frac{- 23 \pm 33}{2} = - 28;5;\ \]

\[1)\ y_{1} = - 28,\ \ x_{1} = 70;\]

\[2)\ y_{2} = 5,\ \ x_{2} = 4.\]

\[\textbf{в)}\ \left\{ \begin{matrix} x + y = 14\ \ \\ \frac{x}{y} + \frac{y}{x} = 2\frac{1}{12} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 12x^{2} - 12y^{2} = 25xy \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 49y^{2} - 686y + 2352 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y² - 14y + 48 = 0 \\ \end{matrix} \right.\ \]

\[1)\ y_{1} = 8,\ \ x_{1} = 6;\]

\[2)\ y_{2} = 6,\ \ x_{2} = 8.\]

\[\textbf{г)}\ \left\{ \begin{matrix} x - y = 2\ \\ \frac{x}{y} - \frac{y}{x} = \frac{5}{6} \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \\ 6x^{2} - 6y^{2} = 5xy \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y² - 14y - 24 = 0 \\ \end{matrix} \right.\ \]

\[D = 7^{2} + 5 \cdot 24 = 169\]

\[y_{1,2} = \frac{7 \pm 13}{5} = 4;\ - 1,2;\]

\[1)\ y_{1} = 4,\ \ x_{1} = 6;\]

\[2)\ y_{2} = - 1,2,\ \ x_{2} = 0,8.\]

\[Ответ:а)\ (1,5;\ - 2);(10;15);\ \ \]

\[\textbf{б)}\ \ (70;\ - 28);\ \ (4;5);\]

\[\textbf{в)}\ (6;8);\ \ (8;6);\ \ \]

\[\textbf{г)}\ \ (6;4);\ \ (0,8;\ - 1,2).\]

\[\boxed{\text{533.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ b_{n + 1} = b_{n} + 3;\ \ \ b_{1} = 10:\ \]

\[b_{2} = b_{1} + 3 = 10 + 3 = 13\]

\[b_{3} = b_{2} + 3 = 13 + 3 = 16\]

\[b_{4} = b_{3} + 3 = 16 + 3 = 19\]

\[b_{5} = b_{4} + 3 = 19 + 3 = 22.\]

\[\textbf{б)}\ b_{n + 1} = \frac{b_{n}}{2};\ \ \ b_{1} = 40:\ \]

\[b_{2} = \frac{b_{1}}{2} = \frac{40}{2} = 20\]

\[b_{3} = \frac{b_{2}}{2} = \frac{20}{2} = 10\]

\[b_{4} = \frac{b_{3}}{2} = 5\]

\[b_{5} = \frac{b_{4}}{2} = 2,5.\]