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МатематикаРешебник по алгебре 9 класс Макарычев Задание 587
Задание 587
\[\boxed{\text{587}\text{\ (587)}\text{.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[x_{1} = 5;\ \ x_{9} = 1 \Longrightarrow d = \frac{x_{9} - x_{1}}{8} =\]
\[= - 0,5.\]
\[x_{1} = 5\ \ \]
\[x_{2} = 4,5\text{\ \ }\]
\[x_{3} = 4\ \ \]
\[x_{4} = 3,5\ \]
\[x_{5} = 3\]
\[x_{6} = 2,5\ \ \]
\[x_{7} = 2\ \]
\[x_{8} = 1,5\]
\[x_{9} = 1.\]
\(\boxed{\text{587.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)
\[\textbf{а)}\ \left\{ \begin{matrix} 9x^{2} + 9y^{2} = 13 \\ 3xy = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{9 \cdot 4}{9y^{2}} + 9y^{2} = 13 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4 + 9y^{4} = 13y² \\ \end{matrix} \right.\ \]
\[9y^{4} - 13y^{2} + 4 = 0\]
\[D = 13^{2} - 4 \cdot 9 \cdot 4 =\]
\[= 169 - 144 = 25 \Longrightarrow\]
\[\Longrightarrow y_{1,2} = \frac{13 \pm 5}{18};\]
\[1)\ y² = 1 \Longrightarrow y = \pm 1;\]
\[2)\ y² = \frac{8}{18},\]
\[\ \ y² = \frac{4}{9} \Longrightarrow y = \pm \frac{2}{3}.\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = \frac{2}{3} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = - \frac{2}{3} \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{3} = \frac{2}{3} \\ x_{3} = 1\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{4} = - \frac{2}{3} \\ x_{4} = - 1. \\ \end{matrix} \right.\ \]
\[Ответ:\left( \frac{2}{3};1 \right);\ \ \left( - \frac{2}{3}; - 1 \right);\]
\[\text{\ \ }\left( 1;\frac{2}{3} \right);\ \ \left( - 1;\ - \frac{2}{3} \right).\]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 29 \\ y^{2} - 4x^{2} = 9 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y^{2} = 4x^{2} + 9\ \ \ \ \ \ \ \ \\ x^{2} + 4x^{2} + 9 = 29 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} 5x^{2} = 20\ \ \ \ \ \ \\ y^{2} = 4x^{2} + 9 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x² = 4\ \ \ \ \ \ \ \ \ \ \ \\ y² = 16 + 9 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ y = \pm 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 5 \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{2} = 2\ \ \\ x_{2} = - 5 \\ \end{matrix} \right.\ ,\ \ \]
\[\left\{ \begin{matrix} y_{3} = - 2 \\ x_{3} = 5\ \ \ \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{4} = - 2 \\ x_{4} = - 5. \\ \end{matrix} \right.\ \]
\[Ответ:(5;2);\ \ ( - 5;2);\ \]
\[\ (5;\ - 2);\ \ ( - 5;\ - 2).\]
\[\textbf{в)}\ \left\{ \begin{matrix} 2x^{2} + xy = 6\ \ \ \ \ \ \ \\ 3x^{2} + xy - x = 6 \\ \end{matrix} \right.\ \ \ \ ( - )\]
\[- x^{2} + x = 0\]
\[- x(x - 1) = 0\]
\[x_{1} = 0\ (не\ подходит);\ \ x_{2} = 1.\]
\[2 \cdot 1 + y = 6\]
\[y = 4.\]
\[Ответ:(1;4).\]
\[\textbf{г)}\ \left\{ \begin{matrix} 3x^{2} - 2y^{2} = 25\ \ \ \ \ \ \ \ \ \ \\ x^{2} - y^{2} + y = 5\ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\text{\ \ }\left\{ \begin{matrix} 3x^{2} - 2y^{2} = 25\ \ \ \ \ \ \ \ \ \ \\ 3x^{2} - 3y^{2} + 3y = 15 \\ \end{matrix} \right.\ ( - )\]
\[y^{2} - 3y = 10\]
\[y^{2} - 3y - 10 = 0\]
\[y_{1} + y_{2} = 3;\ \ y_{1} \cdot y_{2} = - 10\]
\[y_{1} = 5;\ \ y_{2} = - 2.\]
\[3x^{2} = 2y^{2} + 25\]
\[y = 5:\]
\[3x^{2} = 2 \cdot 25 + 25\]
\[3x^{2} = 75\]
\[x^{2} = 25\]
\[x = \pm 5.\]
\[y = - 2:\]
\[3x^{2} = 2 \cdot 4 + 25\]
\[3x^{2} = 33\]
\[x^{2} = 11\]
\[x = \pm \sqrt{11}.\]
\[Ответ:(5;5);( - 5;5);\]
\[\left( - \sqrt{11};\ - 2 \right);\left( \sqrt{11}; - 2 \right).\ \]
