ГДЗ по алгебре 9 класс Макарычев Задание 440

\[\boxed{\text{440\ (440).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 16 \\ x - y = 4\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (y + 4)^{2} + y^{2} = 16 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \\ y(y + 4) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 0 \\ x_{1} = 4 \\ \end{matrix} \right.\ \text{\ \ }или\ \ \left\{ \begin{matrix} y_{2} = - 4 \\ x_{2} = 0\ \ \ . \\ \end{matrix} \right.\ \]

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

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

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

\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 1 \\ y_{1} = 2 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \left\{ \begin{matrix} x_{2} = - 1,5 \\ y_{2} = 3,25. \\ \end{matrix} \right.\ \]

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

\[D = 1 + 24 = 25\]

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

\[x_{2} = \frac{- 1 - 5}{4} = - 1,5.\]

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

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

\[Ответ:а)\ (4;0);(0;\ - 4);\ \ \ \]

\[\textbf{б)}\ (1;2);\ \ ( - 1,5;3,25).\]