ГДЗ по алгебре 9 класс Макарычев Задание 441

\[\boxed{\text{441\ (441).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + xy - y^{2} = 11 \\ x - 2y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} + 4y + 1 + 2y^{2} + y - y^{2} = 11 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 2y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y^{2} + 5y - 10 = 0 \\ \end{matrix} \right.\ \]

\[5y^{2} + 5y - 10 = 0\ \ \ \ |\ :5\]

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

\[y_{1} + y_{2} = - 1;\ \ \ y_{2} \cdot y_{2} = - 2\]

\[y_{1} = 1;\ \ \ y_{2} = - 2.\]

\[1)\ y_{1} = 1;\ \ x_{1} = 3;\]

\[2)\ y_{2} = - 2;\ \ x_{2} = - 3.\ \]

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + xy - 3y = 9 \\ 3x + 2y = - 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = - 1,5x - 0,5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 1,5x^{2} - 0,5x + 4,5x + 1,5 - 9 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = - 1,5x - 0,5\ \ \ \ \ \ \ \ \ \ \\ - 0,5x^{2} + 4x - 7,5 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[0,5x^{2} - 4x + 7,5 = 0\ \ \ \ \ \ | \cdot 2\]

\[x^{2} - 8x + 15 = 0\]

\[D_{1} = 16 - 15 = 1\]

\[x_{1} = 4 + 1 = 5;\ \ \ \]

\[x_{2} = 4 - 1 = 3.\]

\[1)\ x_{1} = 5;\ \ y_{1} = - 8;\]

\[2)\ x_{2} = 3;\ \ \ y_{2} = - 5.\]

\[Ответ:а)\ \ (3;1);( - 3;\ - 2);\ \ \]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (5;\ - 8);\ \ (3; - 5).\]