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МатематикаГДЗ по алгебре 9 класс Макарычев Задание 771
Задание 771
\[\boxed{\text{771\ (771).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[C_{8}^{2} = \frac{8!}{2! \cdot 6!} = \frac{7 \cdot 8}{2} = 28\ прямых\ \]
\[можно\ провести\ через\ 8\ точек.\]
\[\boxed{\text{771.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x + xy + y = 11 \\ x - xy + y = 1\ \ \\ \end{matrix} \right.\ ( - )\]
\[\Longrightarrow \left\{ \begin{matrix} 2xy = 10\ \ \ \ \ \ \ \ \ \ \ \ \\ x + xy + y = 11 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = \frac{5}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x + 5 + \frac{5}{x} = 11 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = \frac{5}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x² + 5x + 5 - 11x = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 6x + 5 = 0\]
\[D = 36 - 20 = 16\]
\[x_{1} = \frac{6 + 4}{2} = 5,\]
\[\text{\ \ }x_{2} = \frac{6 - 4}{2} = 1,\]
\[1)\ x_{1} = 5,\ \ y_{1} = \frac{5}{5} = 1,\]
\[2)\ x_{2} = 1,\ \ y_{2} = \frac{5}{1} = 5.\]
\[Ответ:(5;1)\ или\ \ (1;5).\]
\[\textbf{б)}\ \left\{ \begin{matrix} 2x - y - xy = 14\ \\ x + 2y + xy = - 7 \\ \end{matrix} \right.\ ( + )\]
\[\Longrightarrow \left\{ \begin{matrix} 3x + y = 7\ \ \ \ \ \ \ \ \ \ \ \ \\ x + 2y + xy = - 7 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 7 - 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 3x^{2} + 2x + 21 = 0 \\ \end{matrix} \right.\ \]
\[3x² - 2x - 21 = 0\]
\[D = 4 + 252 = 256\]
\[x_{1} = \frac{2 + 16}{6} = 3,\ \ \]
\[x_{2} = \frac{2 - 16}{6} = - \frac{14}{6} = - 2\frac{1}{3},\]
\[1)\ x_{1} = 3,\ \ y_{1} = - 2,\]
\[2)\ x = - 2\frac{1}{3},\ \ y = 14.\]
\[Ответ:(3;\ - 2)\ или\ \ \left( - 2\frac{1}{3};14 \right).\]
\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} + y^{2} = 34 \\ xy = 15\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x + y)^{2} - 2xy = 34 \\ xy = 15\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} xy = 15\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (x + y)^{2} = 34 + 30 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} xy = 15\ \ \ \ \ \ \ \ \ \ \\ (x + y)^{2} = 64 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} xy = 15\ \ \ \ \ \ \\ x + y = \pm 8 \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x + y = 8 \\ xy = 15\ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 8 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 8x - x^{2} - 15 = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 8x + 15 = 0\]
\[D = 64 - 60 = 4\]
\[x_{1} = \frac{8 + 2}{2} = 5,\]
\[\text{\ \ }x_{2} = \frac{8 - 2}{2} = 3,\]
\[x_{1} = 3,\ \ y_{1} = 5;\]
\[x_{2} = 5,\ \ y_{2} = 3.\]
\[2)\ \left\{ \begin{matrix} x + y = - 8 \\ xy = 15\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} xy = 15\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 8x - x^{2} - 15 = 0 \\ \end{matrix} \right.\ \]
\[x^{2} + 8x + 15 = 0\]
\[D = 64 - 60 = 4\]
\[x_{1} = \frac{- 8 + 2}{2} = - 3,\]
\[\text{\ \ }x_{2} = \frac{- 8 - 2}{2} = - 5,\]
\[x_{1} = - 3,\ \ y_{1} = - 5;\]
\[x_{2} = - 5,\ \ y_{2} = - 3.\]
\[Ответ:(3;5),(\ 5;3),\ \]
\[( - 3;\ - 5),\ ( - 5;\ - 3).\]
\[\textbf{г)}\ \left\{ \begin{matrix} x^{2} - y^{2} = 12 \\ xy = 8\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = \frac{8}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{2} - \frac{64}{x^{2}} = 12 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = \frac{8}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{4} - 64 - 12x^{2} = 0 \\ \end{matrix} \right.\ \]
\[x^{2} = t,\ \ t \geq 0,\]
\[t^{2} - 12t - 64 = 0\]
\[D = 144 + 256 = 400\]
\[t_{1} = \frac{12 + 20}{2} = 16,\]
\[t_{2} = \frac{12 - 20}{2} = - 4 \Longrightarrow\]
\[\Longrightarrow не\ подходит\ по\ условию.\]
\[x^{2} = 16,\ \ x = \pm 4,\]
\[1)\ x_{1} = 4,\ \ y_{1} = 2;\]
\[2)\ x_{2} = - 4,\ \ y_{2} = - 2.\]
\[Ответ:(4;2)\ или\ \ ( - 4;\ - 2).\]
