Алгебра
Английский
Биология
География
Геометрия
История
Русский
Обществознание
Физика
Химия
Информатика
Литература
МатематикаРешебник по алгебре 9 класс Макарычев Задание 479
Задание 479
\[\boxed{\text{479\ (479).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} 3x + y + 4 = 0 \\ x^{2} - y^{2} = 2\ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = - 3x - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - ( - 3x - 4)^{2} = 2 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = - 3x - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 9x^{2} - 24x - 16 = 2 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = - 3x - 4\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8x^{2} + 24x + 18 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[4x^{2} + 12x + 9 = 0\]
\[(2x + 3)^{2} = 0\]
\[2x + 3 = 0\]
\[2x = - 3\]
\[x = - 1,5 \Longrightarrow y =\]
\[= - 3 \cdot ( - 1,5) - 4 = 0,5.\]
\[Ответ:( - 1,5;0,5).\]
\[\textbf{б)}\ \left\{ \begin{matrix} y + 3x = 2\ \ \ \ \ \ \ \\ x^{2} - xy = 3,36 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - x(2 - 3x) = 3,36 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2x + 3x^{2} - 3,36 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 2 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4x^{2} - 2x - 3,36 = 0 \\ \end{matrix} \right.\ \]
\[2x^{2} - x - 1,68 = 0\]
\[D = 1 + 4 \cdot 2 \cdot 1,68 = 14,44\]
\[x_{1,2} = \frac{1 \pm 3,8}{4} = 1,2;\ - 0,7.\]
\[\left\{ \begin{matrix} x_{1} = 1,2\ \ \ \\ y_{1} = - 1,6 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x_{2} = - 0,7 \\ y_{2} = 4,1\ \ \ . \\ \end{matrix} \right.\ \]
\[Ответ:а)\ ( - 1,5;0,5);\ \ \]
\[\textbf{б)}\ \ (1,2;\ - 1,6);( - 0,7;4,1).\]
\[\boxed{\text{479.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + xy + y^{2} = 7 \\ x + xy + y = 5\ \ \\ \end{matrix} \right.\ \]
\[Пусть\ xy = a,\ \ x + y = b,\]
\[\text{\ \ }тогда\ x^{2} + xy + y^{2} =\]
\[= (x + y)^{2} - xy = b^{2} - a;\]
\[\left\{ \begin{matrix} b^{2} - a = 7 \\ b + a = 5\ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} a = 5 - b\ \ \ \ \ \ \ \ \ \ \\ b² + b - 12 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\ \left\{ \begin{matrix} b_{1} = - 4 \\ a_{1} = 9\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = - 4 \\ xy = 9\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 4 - y\ \ \ \\ - y^{2} - 4y = 9 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y^{2} + 4y + 9 = 0 \\ x = - 4 - y\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[D = - 4 - 9 < 0 \Longrightarrow корней\ нет;\]
\[2)\ \left\{ \begin{matrix} b_{2} = 3 \\ a_{2} = 2 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = 3 \\ xy = 2\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 1 \\ \end{matrix} \right.\ \ \ или\ \left\{ \begin{matrix} x_{2} = 1 \\ y_{2} = 2. \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + xy + y^{2} = 19 \\ x + xy + y = 1\ \ \ \ \ \\ \end{matrix} \right.\ \]
\[Пусть\ a = xy,\ \ b = x + y,\]
\[\text{\ \ }тогда\ x^{2} + xy + y^{2} =\]
\[= (x + y)^{2} - xy = b^{2} - a;\]
\[\Longrightarrow \left\{ \begin{matrix} b^{2} - a = 19 \\ b + a = 1\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} a = 1 - b\ \ \ \ \ \ \ \ \ \ \ \\ b² + b - 20 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[1)\ \left\{ \begin{matrix} b_{1} = - 5 \\ a_{1} = 6\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = - 5 \\ xy = 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - y - 5\ \ \ \ \ \ \\ y^{2} + 5y + 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = - 3 \\ x_{1} = - 2 \\ \end{matrix} \right.\ \ \ или\ \ \left\{ \begin{matrix} y_{2} = - 2 \\ x_{2} = - 3 \\ \end{matrix} \right.\ ;\]
\[2)\ \left\{ \begin{matrix} b_{2} = 4\ \ \ \\ a_{2} = - 3 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x + y = 4 \\ xy = - 3\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \\ y(4 - y) = - 3 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y² - 4y - 3 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} - 4y - 3 = 0\]
\[D = 4 + 3 = 7\]
\[y_{1,2} = 2 \pm \sqrt{7};\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 + \sqrt{7} \\ x_{1} = 2 - \sqrt{7} \\ \end{matrix} \right.\ \ \ \ или\ \]
\[\ \left\{ \begin{matrix} y_{2} = 2 - \sqrt{7} \\ x_{2} = 2 + \sqrt{7}. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (1;2);(2;1);\ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ ( - 2;\ - 3);( - 3;\ - 2);\]
\[\left( 2 - \sqrt{7};2 + \sqrt{7} \right);\]
\[\left( 2 + \sqrt{7};2 - \sqrt{7} \right).\]
