ГДЗ по алгебре 9 класс Макарычев Задание 438

\[\boxed{\text{438\ (438).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} - 4 = 0 \\ xy = 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x^{2} = 4 \\ xy = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ y = \frac{6}{x}\text{\ \ \ \ } \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 3 \\ \end{matrix} \right.\ \text{\ \ }или\ \]

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

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} - 5x + 6 = 0 \\ y^{2} - 6y + 5 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} (x - 2)(x - 3) = 0 \\ (y - 1)(y - 5) = 0 \\ \end{matrix} \right.\ \]

\[x^{2} - 5x + 6 = 0\]

\[x_{1} + x_{2} = 5;\ \ \ \ x_{1} \cdot x_{2} = 6\]

\[x_{1} = 2;\ \ \ \ \ x_{2} = 3.\]

\[y^{2} - 6y + 5 = 0\]

\[D_{1} = 9 - 5 = 4\]

\[y_{1} = 3 + 2 = 5;\ \ \]

\[\ y_{2} = 3 - 2 = 1.\]

\[\left\{ \begin{matrix} x_{1} = 2 \\ y_{1} = 1 \\ \end{matrix} \right.\ ;\ \ \ \ \left\{ \begin{matrix} x_{2} = 3 \\ y_{2} = 1 \\ \end{matrix} \right.\ ;\ \ \ \ \]

\[\left\{ \begin{matrix} x_{3} = 2 \\ y_{3} = 5 \\ \end{matrix} \right.\ ;\ \ \ \ \left\{ \begin{matrix} x_{4} = 3 \\ y_{4} = 5 \\ \end{matrix} \right.\ .\]

\[Ответ:а)\ (2;3);( - 2;\ - 3);\ \]

\[\ б)\ (2;1);(3;1);(2;5);(3;5).\]