ГДЗ по алгебре 9 класс Макарычев Задание 527

\[\boxed{\text{527\ (527).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

\[\Longrightarrow \left\{ \begin{matrix} x = - 3y - 1\ \ \ \ \ \ \ \ \\ 3y² + 5y - 2 = 0 \\ \end{matrix} \right.\ \]

\[D = 25 + 4 \cdot 3 \cdot 2 = 49\]

\[y_{1,2} = \frac{- 5 \pm 7}{6} = \frac{1}{3};\ - 2;\]

\[1)\ y_{1} = \frac{1}{3},\ \ x_{1} = - 2;\]

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

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

\[\Longrightarrow \left\{ \begin{matrix} y = 2x - 1\ \ \ \\ 2x^{2} - 6x = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

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

\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 0\ \ \\ y_{1} = - 1 \\ \end{matrix} \right.\ или\ \left\{ \begin{matrix} x_{2} = 3 \\ y_{2} = 5. \\ \end{matrix} \right.\ \]

\[\textbf{в)}\ \left\{ \begin{matrix} 2x + y - 11 = 0\ \ \ \ \ \ \ \ \ \ \\ 2x + 5y - y^{2} - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 11 - 2x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 4x^{2} + 36x - 72 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y = 11 - 2x\ \ \ \ \ \ \\ x^{2} - 9x + 18 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x_{1} = 6\ \ \ \\ y_{1} = - 1 \\ \end{matrix} \right.\ \ \ или\ \ \left\{ \begin{matrix} x_{2} = 3\ \\ y_{2} = 5. \\ \end{matrix} \right.\ \]

\[\textbf{г)}\ \left\{ \begin{matrix} 2x^{2} - 3y^{2} - 5x - 2y = 26 \\ x - y = 4\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y + 4\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 9y + 14 = 0 \\ \end{matrix} \Longrightarrow \right.\ \]

\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 7\ \ \\ x_{1} = 11 \\ \end{matrix} \right.\ \ \ или\ \ \ \left\{ \begin{matrix} y_{2} = 2\ \\ x_{2} = 6. \\ \end{matrix} \right.\ \]

\[\textbf{д)}\ \left\{ \begin{matrix} 4x^{2} - 9y^{2} + x - 40y = 19 \\ 2x - 3y = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = 1,5y + 2,5\ \ \\ - 8,5y + 8,5 = 0 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = 1\ \\ x = 4. \\ \end{matrix} \right.\ \]

\[\textbf{е)}\ \left\{ \begin{matrix} 3x^{2} + y^{2} + 8x + 13y = 5 \\ x - y + 2 = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = y - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y² + 9y - 9 = 0 \\ \end{matrix} \right.\ \]

\[4y^{2} + 9y - 9 = 0\]

\[D = 81 + 4 \cdot 4 \cdot 9 = 225\]

\[y_{1,2} = \frac{- 9 \pm 15}{8} = - 3;\frac{3}{4};\ \]

\[1)\ y_{1} = - 3,\ \ x_{1} = - 5;\]

\[2)\ y_{2} = \frac{3}{4},\ \ x_{2} = - 1\frac{1}{4}.\]

\[Ответ:а)\ (5;\ - 2);\left( - 2;\frac{1}{3} \right);\ \ \]

\[\textbf{б)}\ (0;\ - 1);(3;5);\]

\[\textbf{в)}\ (6;\ - 1);(3;5);\ \ \]

\[\textbf{г)}\ \ (6;2);(11;7);\]

\[\textbf{д)}\ (4;1);\ \ \ \]

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