Решебник по алгебре 9 класс Макарычев Задание 587

\[\boxed{\text{587}\text{\ (587)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[x_{1} = 5;\ \ x_{9} = 1 \Longrightarrow d = \frac{x_{9} - x_{1}}{8} =\]

\[= - 0,5.\]

\[x_{1} = 5\ \ \]

\[x_{2} = 4,5\text{\ \ }\]

\[x_{3} = 4\ \ \]

\[x_{4} = 3,5\ \]

\[x_{5} = 3\]

\[x_{6} = 2,5\ \ \]

\[x_{7} = 2\ \]

\[x_{8} = 1,5\]

\[x_{9} = 1.\]

\(\boxed{\text{587.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)

\[\textbf{а)}\ \left\{ \begin{matrix} 9x^{2} + 9y^{2} = 13 \\ 3xy = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{9 \cdot 4}{9y^{2}} + 9y^{2} = 13 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4 + 9y^{4} = 13y² \\ \end{matrix} \right.\ \]

\[9y^{4} - 13y^{2} + 4 = 0\]

\[D = 13^{2} - 4 \cdot 9 \cdot 4 =\]

\[= 169 - 144 = 25 \Longrightarrow\]

\[\Longrightarrow y_{1,2} = \frac{13 \pm 5}{18};\]

\[1)\ y² = 1 \Longrightarrow y = \pm 1;\]

\[2)\ y² = \frac{8}{18},\]

\[\ \ y² = \frac{4}{9} \Longrightarrow y = \pm \frac{2}{3}.\]

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

\[\Longrightarrow \left\{ \begin{matrix} y_{3} = \frac{2}{3} \\ x_{3} = 1\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{4} = - \frac{2}{3} \\ x_{4} = - 1. \\ \end{matrix} \right.\ \]

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

\[\text{\ \ }\left( 1;\frac{2}{3} \right);\ \ \left( - 1;\ - \frac{2}{3} \right).\]

\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 29 \\ y^{2} - 4x^{2} = 9 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} y^{2} = 4x^{2} + 9\ \ \ \ \ \ \ \ \\ x^{2} + 4x^{2} + 9 = 29 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} 5x^{2} = 20\ \ \ \ \ \ \\ y^{2} = 4x^{2} + 9 \\ \end{matrix} \right.\ \Longrightarrow\]

\[\Longrightarrow \left\{ \begin{matrix} x² = 4\ \ \ \ \ \ \ \ \ \ \ \\ y² = 16 + 9 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ y = \pm 5 \\ \end{matrix} \right.\ \Longrightarrow\]

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

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

\[Ответ:(5;2);\ \ ( - 5;2);\ \]

\[\ (5;\ - 2);\ \ ( - 5;\ - 2).\]

\[\textbf{в)}\ \left\{ \begin{matrix} 2x^{2} + xy = 6\ \ \ \ \ \ \ \\ 3x^{2} + xy - x = 6 \\ \end{matrix} \right.\ \ \ \ ( - )\]

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

\[- x(x - 1) = 0\]

\[x_{1} = 0\ (не\ подходит);\ \ x_{2} = 1.\]

\[2 \cdot 1 + y = 6\]

\[y = 4.\]

\[Ответ:(1;4).\]

\[\textbf{г)}\ \left\{ \begin{matrix} 3x^{2} - 2y^{2} = 25\ \ \ \ \ \ \ \ \ \ \\ x^{2} - y^{2} + y = 5\ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} 3x^{2} - 2y^{2} = 25\ \ \ \ \ \ \ \ \ \ \\ 3x^{2} - 3y^{2} + 3y = 15 \\ \end{matrix} \right.\ ( - )\]

\[y^{2} - 3y = 10\]

\[y^{2} - 3y - 10 = 0\]

\[y_{1} + y_{2} = 3;\ \ y_{1} \cdot y_{2} = - 10\]

\[y_{1} = 5;\ \ y_{2} = - 2.\]

\[3x^{2} = 2y^{2} + 25\]

\[y = 5:\]

\[3x^{2} = 2 \cdot 25 + 25\]

\[3x^{2} = 75\]

\[x^{2} = 25\]

\[x = \pm 5.\]

\[y = - 2:\]

\[3x^{2} = 2 \cdot 4 + 25\]

\[3x^{2} = 33\]

\[x^{2} = 11\]

\[x = \pm \sqrt{11}.\]

\[Ответ:(5;5);( - 5;5);\]

\[\left( - \sqrt{11};\ - 2 \right);\left( \sqrt{11}; - 2 \right).\ \]