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МатематикаРешебник по алгебре 9 класс Макарычев Задание 498
Задание 498
\[\boxed{\text{498\ (498).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x \geq 2 \\ y \geq 1 \\ \end{matrix} \right.\ \ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x < - 1 \\ y > 0\ \ \ \\ \end{matrix} \right.\ \]
\[\textbf{в)}\ \left\{ \begin{matrix} x + 2 \geq 0 \\ y - 3 \leq 0 \\ \end{matrix} \right.\ \]
\[\boxed{\text{498.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} \frac{1}{x} + \frac{1}{y} = \frac{1}{6}\text{\ \ } \\ 2x - y = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 6y + 6x = xy \\ 2x - y = 5\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 6 \cdot (2x - 5) + 6x = x(2x - 5) \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x² - 23x + 30 = 0 \\ \end{matrix} \right.\ \]
\[D = 23^{2} - 4 \cdot 2 \cdot 30 = 289\]
\[x_{1,2} = \frac{23 \pm 17}{4} = 10;\frac{3}{2};\ \]
\[1)\ x_{1} = 10,\ \ y_{1} = 15;\]
\[2)\ x_{2} = 1,5;\ \ \ \ y_{2} = - 2.\]
\[\textbf{б)}\ \left\{ \begin{matrix} \frac{1}{x} - \frac{1}{y} = \frac{1}{20} \\ x + 2y = 14 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 20y - 20x = xy \\ x = 14 - 2y\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 20y - 280 + 40y = 14y - 2y^{2} \\ x = 14 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} y² + 23y - 140 = 0 \\ x = 14 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[D = 23^{2} + 4 \cdot 140 = 1089 = 33^{2}\]
\[y_{1,2} = \frac{- 23 \pm 33}{2} = - 28;5;\ \]
\[1)\ y_{1} = - 28,\ \ x_{1} = 70;\]
\[2)\ y_{2} = 5,\ \ x_{2} = 4.\]
\[\textbf{в)}\ \left\{ \begin{matrix} x + y = 14\ \ \\ \frac{x}{y} + \frac{y}{x} = 2\frac{1}{12} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 12x^{2} - 12y^{2} = 25xy \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 49y^{2} - 686y + 2352 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 14 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y² - 14y + 48 = 0 \\ \end{matrix} \right.\ \]
\[1)\ y_{1} = 8,\ \ x_{1} = 6;\]
\[2)\ y_{2} = 6,\ \ x_{2} = 8.\]
\[\textbf{г)}\ \left\{ \begin{matrix} x - y = 2\ \\ \frac{x}{y} - \frac{y}{x} = \frac{5}{6} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \\ 6x^{2} - 6y^{2} = 5xy \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y² - 14y - 24 = 0 \\ \end{matrix} \right.\ \]
\[D = 7^{2} + 5 \cdot 24 = 169\]
\[y_{1,2} = \frac{7 \pm 13}{5} = 4;\ - 1,2;\]
\[1)\ y_{1} = 4,\ \ x_{1} = 6;\]
\[2)\ y_{2} = - 1,2,\ \ x_{2} = 0,8.\]
\[Ответ:а)\ (1,5;\ - 2);(10;15);\ \]
\[\ б)\ \ (70;\ - 28);\ \ (4;5);\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ в)\ (6;8);\ \ (8;6);\ \]
\[\ г)\ \ (6;4);\ \ (0,8;\ - 1,2).\]
