Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре 9 класс Макарычев Задание 514
Задание 514
\[\boxed{\text{514\ (514).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + xy + y^{2} = 7 \\ x + xy + y = 5\ \ \\ \end{matrix} \right.\ \]
\[Пусть\ xy = a,\ \ x + y = b,\]
\[тогда\]
\[x^{2} + xy + y^{2} = (x + y)^{2} - xy =\]
\[= b^{2} - a;\]
\[\left\{ \begin{matrix} b^{2} - a = 7 \\ b + a = 5\ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} a = 5 - b\ \ \ \ \ \ \ \ \ \ \\ b² + b - 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\ \left\{ \begin{matrix} b_{1} = - 4 \\ a_{1} = 9\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = - 4 \\ xy = 9\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 4 - y\ \ \ \\ - y^{2} - 4y = 9 \\ \end{matrix} \right.\ \Longrightarrow\]
\[= \left\{ \begin{matrix} y^{2} + 4y + 9 = 0 \\ x = - 4 - y\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[D = - 4 - 9 < 0 \Longrightarrow корней\ нет;\]
\[2)\ \left\{ \begin{matrix} b_{2} = 3 \\ a_{2} = 2 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = 3 \\ xy = 2\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 1 \\ \end{matrix} \right.\ \ \ или\ \left\{ \begin{matrix} x_{2} = 1 \\ y_{2} = 2. \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + xy + y^{2} = 19 \\ x + xy + y = 1\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Пусть\ a = xy,\ \ b = x + y,\]
\[тогда\ \]
\[x^{2} + xy + y^{2} = (x + y)^{2} - xy =\]
\[= b^{2} - a;\]
\[\Longrightarrow \left\{ \begin{matrix} b^{2} - a = 19 \\ b + a = 1\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} a = 1 - b\ \ \ \ \ \ \ \ \ \ \ \\ b² + b - 20 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\ \left\{ \begin{matrix} b_{1} = - 5 \\ a_{1} = 6\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = - 5 \\ xy = 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - y - 5\ \ \ \ \ \ \\ y^{2} + 5y + 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = - 3 \\ x_{1} = - 2 \\ \end{matrix} \right.\ \ \ или\ \ \left\{ \begin{matrix} y_{2} = - 2 \\ x_{2} = - 3 \\ \end{matrix} \right.\ ;\]
\[2)\ \left\{ \begin{matrix} b_{2} = 4\ \ \ \\ a_{2} = - 3 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = 4 \\ xy = - 3\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \\ y(4 - y) = - 3 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y² - 4y - 3 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} - 4y - 3 = 0\]
\[D = 4 + 3 = 7\]
\[y_{1,2} = 2 \pm \sqrt{7};\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 + \sqrt{7} \\ x_{1} = 2 - \sqrt{7} \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[или\ \ \left\{ \begin{matrix} y_{2} = 2 - \sqrt{7} \\ x_{2} = 2 + \sqrt{7}. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (1;2);(2;1);\ \]
\[\textbf{б)}\ ( - 2;\ - 3);( - 3;\ - 2);\]
\[\left( 2 - \sqrt{7};2 + \sqrt{7} \right);\]
\[\left( 2 + \sqrt{7};2 - \sqrt{7} \right).\]
\[\boxed{\text{514.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[Пусть\ \text{x\ }\frac{км}{ч}\ и\ \text{y\ }\frac{км}{ч} -\]
\[скорости\ туристов.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} \frac{9y}{x} = \frac{8x - 6y}{y} \\ 8x - 9y = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} \frac{24y}{4 + 3y} = \frac{12 + 3y}{y} \\ x = \frac{3 \cdot (4 + 3y)}{8}\text{\ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 8y^{2} = 16 + 4y + 12y + 3y^{2} \\ x = \frac{12 + 9y}{8}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 5y² - 16y - 16 = 0 \\ x = \frac{12 + 9y}{8}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[D = 8^{2} + 5 \cdot 16 = 144\]
\[y_{1,2} = \frac{8 \pm 12}{5} = 4;\ - \frac{4}{5};\ \]
\[\left\{ \begin{matrix} y = 4 \\ x = 6 \\ \end{matrix} \right.\ .\]
\[Ответ:скорость\ первого\ \]
\[6\ \frac{км}{ч},\ скорость\ \]
\[второго\ 4\ \frac{км}{ч}.\]
