Решебник по алгебре 9 класс Мерзляк Задание 456

Авторы:
Год:2023
Тип:учебник
Серия:Алгоритм успеха

Задание 456

\[\boxed{\text{456\ (456).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left\{ \begin{matrix} x^{2} - xy + y^{2} = 63 \\ y - x = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x^{2} - xy + y^{2} = 63 \\ y = 3 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x^{2} - x(3 + x) + (3 + x)^{2} -\]

\[- 63 = 0\]

\[x^{2} - 3x - x^{2} + 9 + 6x + x^{2} -\]

\[- 63 = 0\]

\[x^{2} + 3x - 54 = 0\]

\[x_{1} + x_{2} = - 3,\ \ x_{1} = - 9\]

\[x_{1}x_{2} = - 54,\ \ \ \ \ \ \ \ \ \ \ \ x_{2} = 6\]

\[\left\{ \begin{matrix} x = - 9 \\ y = - 6 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \ \left\{ \begin{matrix} x = 6 \\ y = 9 \\ \end{matrix} \right.\ \]

\[Ответ:( - 9;\ - 6);\ (6;9).\]

\[2)\ \left\{ \begin{matrix} x + 2y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + xy + 2y^{2} = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 1 - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (1 - 2y)^{2} + (1 - 2y)y + 2y^{2} - 1 = 0 \\ \end{matrix} \right.\ \]

\[1 - 4y + 4y^{2} + y - 2y^{2} +\]

\[+ 2y^{2} - 1 = 0\]

\[4y^{2} - 3y = 0\]

\[y(4y - 3) = 0\]

\[\left\{ \begin{matrix} x = 1 \\ y = 0 \\ \end{matrix} \right.\ \ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x = - 0,5 \\ y = \frac{3}{4}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[Ответ:(1;0);\ \ ( - 0,5;\ 0,75).\]

\[3)\ \left\{ \begin{matrix} (x - 1)(y - 2) = 2 \\ x + y = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\ \left\{ \begin{matrix} (x - 1)(y - 2) = 2 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[(6 - y - 1)(y - 2) - 2 = 0\]

\[(5 - y)(y - 2) - 2 = 0\]

\[7y - y^{2} - 10 - 2 = 0\]

\[- y^{2} + 7y - 12 = 0\]

\[x_{1} + x_{2} = 7,\ \ x_{1} = 3\]

\[x_{1}x_{2} = 12,\ \ \ \ \ \ \ \ \ \ \ \ x_{2} = 4\]

\[\left\{ \begin{matrix} x = 3 \\ y = 3 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \left\{ \begin{matrix} x = 2 \\ y = 4 \\ \end{matrix} \right.\ \]

\[Ответ:(3;3);\ (2;4).\]

\[3)\ \left\{ \begin{matrix} 5x - 2y = 3\ \ | \cdot 4 \\ 3x^{2} - 8y = - 5\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 20x - 8y = 12 \\ 3x^{2} - 8y = - 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 20x - 8y = 12 \\ 3x^{2} + 5 = 8y\ \ \\ \end{matrix} \right.\ \]

\[20x - 3x^{2} - 5 - 12 = 0\]

\[- 3x^{2} + 20x - 17 = 0\]

\[D = 400 - 204 = 196\]

\[x_{1} = \frac{- 20 + 14}{- 6} = 1\]

\[x_{2} = \frac{- 20 - 14}{- 6} = \frac{17}{3}\]

\[\left\{ \begin{matrix} x = 1\ \ \ \ \ \ \ \ \ \\ 3 + 5 = 8y \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \]

\[\left\{ \begin{matrix} x = \frac{17}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{3 \cdot 17^{2}}{9} + 5 = 8y \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} x = 1 \\ y = 1 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \left\{ \begin{matrix} x = \frac{17}{3} = 5\frac{2}{3}\text{\ \ \ } \\ y = \frac{38}{3} = 12\frac{2}{3} \\ \end{matrix} \right.\ \]

\[Ответ:(1;\ 1);\ \left( 5\frac{2}{3};12\frac{2}{3} \right)\text{.\ }\]

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