ГДЗ по алгебре 9 класс Макарычев Задание 431

\[\boxed{\text{431\ (431).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \left\{ \begin{matrix} x - y = 3 \\ xy = 2\ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]

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

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

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

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

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

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

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

\[\textbf{б)}\ \left\{ \begin{matrix} x + y = 2,5 \\ xy = 1,5\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

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

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

\[y^{2} - 2,5y + 1,5 = 0\ \ \ \ \ \]

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

\[y_{1} = 1;\ \ \ \ \ \ \ y_{2} = 1,5;\]

\[1)\ \left\{ \begin{matrix} y_{1} = 1\ \ \ \ \\ x_{1} = 1,5 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} y_{2} = 1,5 \\ x_{2} = 1\ \ \ \ \\ \end{matrix}. \right.\ \]

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

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

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

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

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

\[1)\ \left\{ \begin{matrix} y_{1} = 0\ \ \ \\ x_{1} = - 1 \\ \end{matrix} \right.\ \]

\[2)\ \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = 0\ \ \ \\ \end{matrix} \right.\ .\]

\[\textbf{г)}\ \left\{ \begin{matrix} x - y = 2\ \ \ \ \ \ \ \\ x^{2} - y^{2} = 17 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (y + 2)^{2} - y^{2} = 17 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 4y + 4 - y^{2} - 17 = 0 \\ \end{matrix} \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 2 \\ 4y = 13\ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = 3\frac{1}{4} \\ x = 5\frac{1}{4}. \\ \end{matrix} \right.\ \]

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

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (1,5;1);(1;1,5);\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ в)\ ( - 1;0);(0;\ - 1);\]

\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ г)\ \left( 5\frac{1}{4};3\frac{1}{4} \right).\]