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МатематикаРешебник по алгебре 9 класс Макарычев Задание 621
Задание 621
\[\boxed{\text{621\ (621).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} 9x² + 9y² = 13 \\ 3xy = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{9 \cdot 4}{9y²} + 9y² = 13 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \frac{2}{3y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4 + 9y^{4} = 13y² \\ \end{matrix} \right.\ \]
\[9y^{4} - 13y^{2} + 4 = 0\]
\[D = 13^{2} - 4 \cdot 9 \cdot 4 = 169 - 144 = 25 \Longrightarrow y_{1,2} = \frac{13 \pm 5}{18};\]
\[1)\ y² = 1 \Longrightarrow y = \pm 1;\]
\[2)\ y² = \frac{8}{18},\ \ y² = \frac{4}{9} \Longrightarrow y = \pm \frac{2}{3}.\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 1 \\ x_{1} = \frac{2}{3} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = - \frac{2}{3} \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{3} = \frac{2}{3} \\ x_{3} = 1\ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y_{4} = - \frac{2}{3} \\ x_{4} = - 1. \\ \end{matrix} \right.\ \]
\[Ответ:\left( \frac{2}{3};1 \right);\ \ \left( - \frac{2}{3}; - 1 \right);\ \ \left( 1;\frac{2}{3} \right);\ \ \left( - 1;\ - \frac{2}{3} \right).\]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 29 \\ y^{2} - 4x^{2} = 9 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y^{2} = 4x^{2} + 9\ \ \ \ \ \ \ \ \\ x^{2} + 4x^{2} + 9 = 29 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} 5x^{2} = 20\ \ \ \ \ \ \\ y^{2} = 4x^{2} + 9 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x² = 4\ \ \ \ \ \ \ \ \ \ \ \\ y² = 16 + 9 \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = \pm 2 \\ y = \pm 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 5 \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{2} = 2\ \ \\ x_{2} = - 5 \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{3} = - 2 \\ x_{3} = 5\ \ \ \\ \end{matrix} \right.\ ,\ \ \left\{ \begin{matrix} y_{4} = - 2 \\ x_{4} = - 5. \\ \end{matrix} \right.\ \]
\[Ответ:(5;2);\ \ ( - 5;2);\ \ (5;\ - 2);\ \ ( - 5;\ - 2).\]
\[\boxed{\text{621.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ x_{5} = x_{1}q^{4} = 1\frac{1}{9},\ \ q = \frac{1}{3}:\ \ \]
\[\frac{10}{9} = x_{1} \cdot \frac{1}{81};\ \ \ \ \ \ x_{1} = 90\ \ \ \ \]
\[S_{5} = x_{1} \cdot \frac{q^{5} - 1}{q - 1} =\]
\[= 90 \cdot \frac{\frac{1}{243} - 1\ }{\frac{1}{3} - 1\ } = \frac{90 \cdot 242 \cdot 3}{2 \cdot 243} =\]
\[= 134\frac{4}{9}.\]
\[\textbf{б)}\ x_{4} = x_{1} \cdot q^{3} = 121,5,\]
\[\ \ q = - 3:\ \ \]
\[121,5 = x_{1} \cdot ( - 3)^{3};\ \ \ \]
\[\text{\ \ }x_{1} = - 4,5\]
\[S_{5} = - 4,5 \cdot \frac{( - 3)^{5} - 1}{- 3 - 1} =\]
\[= - \frac{244 \cdot 9}{4 \cdot 2} = - 274,5.\]
\[Ответ:а)\ 134\frac{4}{9};\ \ б) - 274,5.\]
