ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 703

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\[\textbf{а)}\ \left\{ \begin{matrix} x = 3 - y\ \ \ \ \\ y^{2} - x = 39 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 3 + y - 39 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + y - 42 = 0 \\ \end{matrix} \right.\ \]

\[y^{2} + y - 42 = 0\]

\[y_{1} + y_{2} = - 1;\ \ y_{1} \cdot y_{2} = - 42\]

\[y_{1} = - 7;\ \ y_{2} = 6;\]

\[1)\ \left\{ \begin{matrix} y_{1} = - 7 \\ x_{1} = 10 \\ \end{matrix} \right.\ ;\ \ \ \ \ \ \ 2)\ \left\{ \begin{matrix} y_{2} = 6\ \ \ \\ x_{2} = - 3 \\ \end{matrix} \right.\ \]

\[Ответ:(10; - 7);( - 3;6).\]

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

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

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

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

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

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

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

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

\[Ответ:( - 1;0);( - 2; - 1).\]

\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} + y = 14 \\ y - x = 8\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = x + 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + x + 8 - 14 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = x + 8\ \ \ \ \ \ \ \ \ \\ x^{2} + x - 6 = 0 \\ \end{matrix} \right.\ \]

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

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

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

\[1)\ \left\{ \begin{matrix} x_{1} = 2\ \ \\ y_{1} = 10 \\ \end{matrix} \right.\ ;\ \ \ \ 2)\ \left\{ \begin{matrix} x_{2} = - 3 \\ y_{2} = 5\ \ \ \\ \end{matrix} \right.\ .\]

\[Ответ:( - 3;5);\ \ (2;10).\]

\[\textbf{г)}\ \left\{ \begin{matrix} x + y = 4\ \ \\ y + xy = 6 \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y + 4y - y^{2} - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 5y + 6 = 0 \\ \end{matrix} \right.\ \]

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

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

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

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

\[Ответ:(2;2);\ \ (1;3).\]