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МатематикаРешебник по алгебре 9 класс Макарычев Задание 429
Задание 429
\[\boxed{\text{429\ (429).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} y^{2} - x = - 1 \\ x = y + 3\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y^{2} - y - 3 = - 1 \\ x = y + 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y^{2} - y - 2 = 0 \\ x = y + 3\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[y^{2} - y - 2 = 0\]
\[y_{1} + y_{2} = 1;\ \ y_{1} \cdot y_{2} = - 2\]
\[y_{1} = - 1;\ \ \ \ \ \ \ \ \ \ y_{2} = 2.\]
\[\left\{ \begin{matrix} y_{1} = - 1 \\ x_{1} = 2\ \ \ \\ \end{matrix} \right.\ \ \ \ \ или\ \ \left\{ \begin{matrix} y_{2} = 2 \\ x_{2} = 5 \\ \end{matrix} \right.\ \]
\[Ответ:\ \ (2;\ - 1);(5;2).\]
\[\textbf{б)}\ \left\{ \begin{matrix} y = x - 1\ \ \ \ \ \\ x^{2} - 2y = 26 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2 \cdot (x - 1) = 26 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 2x - 24 = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 2x - 24 = 0\]
\[D_{1} = 1 + 24 = 25\]
\[x_{1} = 1 + 5 = ;\ \ \ x_{2} = 1 - 5 = - 4.\]
\[\ \left\{ \begin{matrix} x_{1} = 6 \\ y_{1} = 5 \\ \end{matrix} \right.\ \ \ \ или\ \ \left\{ \begin{matrix} x_{2} = - 4 \\ y_{2} = - 5 \\ \end{matrix} \right.\ .\]
\[Ответ:(6;5);( - 4;\ - 5).\]
\[\textbf{в)}\ \left\{ \begin{matrix} xy + x = - 4 \\ x - y = 6\ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x - 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x(x - 6) + x = - 4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x - 6\ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 5x + 4 = 0 \\ \end{matrix} \right.\ \]
\[x^{2} - 5x + 4 = 0\]
\[x_{1} + x_{2} = 5;\ \ \ x_{1} \cdot x_{2} = 4\]
\[x_{1} = 1;\ \ \ \ x_{2} = 4.\]
\[\left\{ \begin{matrix} x_{1} = 1\ \ \ \\ y_{1} = - 5 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} x_{2} = 4\ \ \ \ \\ y_{2} = - 2 \\ \end{matrix} \right.\ .\]
\[Ответ:(1;\ - 5);(4;\ - 2).\]
\[\textbf{г)}\ \left\{ \begin{matrix} x + y = 9\ \ \ \\ y^{2} + x = 29 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 9 - y\ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 9 - y = 29 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 9 - y\ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - y - 20 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} - y - 20 = 0\]
\[y_{1} + y_{2} = 1;\ \ \ y_{1} \cdot y_{2} = - 20\]
\[y_{1} = 5;\ \ \ y_{2} = - 4.\]
\[\left\{ \begin{matrix} y_{1} = 5 \\ x_{1} = 4 \\ \end{matrix} \right.\ \ \ или\ \ \left\{ \begin{matrix} y_{2} = - 4 \\ x_{2} = 13. \\ \end{matrix} \right.\ \]
\[Ответ:(4;5);\ \ (13;\ - 4).\]
\(\boxed{\text{429.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)
\[Пусть\ катеты\ треугольника\]
\[\ равны\ x\ и\ y\ см.\ \]
\[По\ теореме\ Пифагора:\]
\[x^{2} + y^{2} = 13^{2}.\]
\[Если\ один\ и\ катетов\ равен\ \]
\[(x + 4)\ см,\ то\ гипотенуза\ \]
\[станет\ равна:\]
\[(x + 4)^{2} + y^{2} = (13 + 2)^{2}.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} x^{2} + y^{2} = 169\ \ \ \ \ \ \ \ \ \ \ \\ (x + 4)^{2} + y^{2} = 225 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\left\{ \begin{matrix} y^{2} = 169 - x^{2} \\ 8x = 40\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} x = 5\ \ \ \ \ \ \ \\ y^{2} = 144 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 5\ \ \ \\ y = 12 \\ \end{matrix}. \right.\ \]
\[Ответ:5\ см\ и\ 12\ см.\]
