Алгебра
Английский
Биология
География
Геометрия
Русский
История
Обществознание
Физика
Химия
Информатика
Литература
МатематикаГДЗ по алгебре 9 класс Макарычев Задание 508
Задание 508
\[\boxed{\text{508\ (508).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + xy - 2y^{2} - x + y = 0 \\ x^{2} + y^{2} = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x + 2y - 1)(x - y) = 0 \\ x² + y² = 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x + 2y - 1 = 0 \\ x^{2} + y^{2} = 8\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 1 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ (1 - 2y)^{2} + y^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 1 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \\ 5y² - 4y - 7 = 0 \\ \end{matrix} \right.\ \]
\[5y^{2} - 4y - 7 = 0\]
\[D = 4 + 35 = 39\]
\[y_{1,2} = \frac{2 \pm \sqrt{39}}{5};\]
\[\left\{ \begin{matrix} y_{1} = \frac{2 + \sqrt{39}}{5} \\ x_{1} = \frac{1 - 2\sqrt{39}}{5}\ \\ \end{matrix} \right.\ \text{\ \ }\]
\[или\ \ \left\{ \begin{matrix} y_{2} = \frac{2 - \sqrt{39}}{5} \\ x_{2} = \frac{1 + 2\sqrt{39}}{5}. \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x - y = 0\ \ \ \\ x^{2} + y^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y\ \ \ \ \ \ \ \ \ \ \\ x^{2} + x^{2} = 8 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = y \\ x^{2} = 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 2 \\ \end{matrix} \right.\ \text{\ \ }или\left\{ \begin{matrix} x_{2} = - 2 \\ y_{2} = - 2. \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} - 6xy + 5y^{2} - x + 5y = 0 \\ x^{2} - 20y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x - y - 1)(x - 5y) = 0 \\ x² - 20y^{2} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x - y - 1 = 0 \\ x^{2} - {20y}^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 2y + 1 - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 19y² - 2y + 4 = 0 \\ \end{matrix} \right.\ \]
\[19y^{2} - 2y + 4 = 0\]
\[D = 1 - 19 \cdot 4 < 0 \Longrightarrow корней\ \]
\[нет;\]
\[2)\ \left\{ \begin{matrix} x - 5y = 0\ \ \ \ \ \\ x^{2} - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 25y^{2} - 20y^{2} = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 5y \\ y^{2} = 1\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = \pm 1 \\ x = 5y \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = 5 \\ \end{matrix} \right.\ \text{\ \ \ }или\left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = - 5. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ ( - 2;\ - 2);(2;2);\]
\[\left( \frac{1 + 2\sqrt{39}}{5};\ \frac{2 - \sqrt{39}}{5} \right);\]
\[\left( \frac{1 - 2\sqrt{39}}{5};\ \frac{2 + \sqrt{39}}{5} \right);\]
\[\ б)\ ( - 5;\ - 1);(5;1).\]
\[\boxed{\text{508.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[Пусть\ x - числитель\ дроби,\ \]
\[а\ y - знаменатель.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} \frac{x + 7}{y^{2}} = \frac{3}{4} \\ \frac{x}{y + 6} = \frac{1}{2} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 4x + 28 = 3y^{2} \\ y + 6 = 2x\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x + 28 = 3 \cdot (2x - 6)^{2} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x + 28 = 12x^{2} - 72x + 108 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 12x^{2} - 76x + 80 = 0 \\ \end{matrix} \Longrightarrow \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3x^{2} - 19x + 20 = 0 \\ \end{matrix} \right.\ \]
\[D = 19^{2} - 4 \cdot 3 \cdot 20 = 121\]
\[x_{1,2} = \frac{19 \pm 11}{6} = 5;\frac{4}{3};\ \]
\[\left\{ \begin{matrix} x_{1} = 5 \\ y_{1} = 4 \\ \end{matrix} \right.\ \ \ \ или\ x_{2} = \frac{4}{3} \Longrightarrow\]
\[\Longrightarrow неподходит\ по\ условию.\]
\[Ответ:\ \frac{5}{4}.\]
