ГДЗ по алгебре 9 класс Мерзляк Задание 455

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Год:2023
Тип:учебник
Серия:Алгоритм успеха

Задание 455

\[\boxed{\text{455\ (455).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\left\{ \begin{matrix} 3x + 4y = 24 \\ xy = 12\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 3x + 4y = 24 \\ x = \frac{12}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} \frac{3 \cdot 12}{y} + 4y = 24 \\ x = \frac{12}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{36}{y} + 4y = 24\ \ |\ :4 \\ x = \frac{12}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{9}{y} + y = 6 \\ x = \frac{12}{y}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\frac{9 + y^{2} - 6y}{y} = 0\]

\[\left\{ \begin{matrix} y^{2} - 6y + 9 = 0 \\ y \neq 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y = 3\ \ \ \\ x = \frac{12}{y} \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

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

\[Ответ:(4;3).\]

\[2)\ \left\{ \begin{matrix} y + 2x = 0\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} - 6y = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} y = - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 4x^{2} + 12x = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\ \left\{ \begin{matrix} 5x^{2} + 12x = 0 \\ y = - 2x\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x(5x + 12) = 0 \\ y = - 2x\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} x = 0 \\ y = 0 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \]

\[\ \left\{ \begin{matrix} x = - \frac{12}{5} = - 2,4 \\ y = 4,8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]

\[Ответ:(0;0);\ \ ( - 2,4;4,8).\]

\[3)\ \left\{ \begin{matrix} x^{2} - xy - y^{2} = 19 \\ x - y = 7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} (7 + y)^{2} - (7 + y)y - y^{2} = 19 \\ x = 7 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[49 + 14y + y^{2} - 7y - y^{2} -\]

\[- y^{2} = 19\]

\[- y^{2} + 7y + 30 = 0\ \]

\[\left\{ \begin{matrix} y_{1} + y_{2} = 7 \\ y_{1}y_{2} = - 30 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y_{1} = 10\ \\ y_{2} = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} x = 17 \\ y = 10 \\ \end{matrix}\ \right.\ \text{\ \ \ }или\ \ \left\{ \begin{matrix} x = 4 \\ y = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:(17;10);\ (\ 4;\ - 3).\]

\[4)\ \left\{ \begin{matrix} x + y = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x - 3)(y + 5) = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (5 - y - 3)(y + 5) = 6 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} (2 - y)(y + 5) = 6 \\ x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} 2y + 10 - y^{2} - 5y - 6 = 0 \\ x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

\[\left\{ \begin{matrix} y_{1} + y_{2} = - 3 \\ y_{1}y_{2} = - 4\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y_{1} = - 4 \\ y_{2} = 1\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 9\ \ \ \ \\ y = - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \left\{ \begin{matrix} x = 4 \\ y = 1 \\ \end{matrix} \right.\ \]

\[Ответ:(9;\ - 4);\ (4;1).\]

\[5)\ \left\{ \begin{matrix} 4y - 3x = 4\ \ \ | \cdot 4 \\ 5x^{2} + 16y = 60\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 16y - 12x = 16 \\ 16y + 5x^{2} = 60 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} 16y = 16 + 12x\ \ \ |\ :4 \\ 5x^{2} + 16y - 60 = 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} 4y = 4 + 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5x^{2} + 16y - 60 = 0 \\ \end{matrix} \right.\ \]

\[5x^{2} + 16 + 12x - 60 = 0\]

\[5x^{2} + 12x - 44 = 0\]

\[D = 144 + 880 = 1024\]

\[x_{1} = \frac{- 12 + 32}{10} = 2,\ \ \]

\[x_{2} = \frac{- 12 - 32}{10} = - 4,4\]

\[\left\{ \begin{matrix} 4y = 4 + 6 \\ x = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \]

\[\ \left\{ \begin{matrix} 4y = 4 - 13,2 \\ x = - 4,4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 2,5 \\ x = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \left\{ \begin{matrix} y = - 2,3 \\ x = - 4,4 \\ \end{matrix} \right.\ \]

\[Ответ:(2;2,5);\ ( - 4,4;\ - 2,3).\]

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

\[\left\{ \begin{matrix} x^{2} + 3xy + y^{2} - x - 2y - 3 = 0 \\ x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

\[- 3 + y - 2y - 3 = 0\]

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

\[- 3 + y - 2y - 3 = 0\]

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

\[\left\{ \begin{matrix} y_{1} + y_{2} = 2 \\ y_{1}y_{2} = - 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y_{1} = 3 \\ y_{2} = - 1 \\ \end{matrix} \right.\ \]

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

\[Ответ:(0;3);\ (4;\ - 1).\]

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