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МатематикаРешебник по алгебре 9 класс Макарычев Задание 430
Задание 430
\[\boxed{\text{430\ (430).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} x = 3 - y\ \ \ \ \\ y^{2} - x = 39 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 3 + y - 39 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + y - 42 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} + y - 42 = 0\]
\[y_{1} + y_{2} = - 1;\ \ \ y_{1} \cdot y_{2} = - 42\]
\[y_{1} = - 7;\ \ y_{2} = 6.\]
\[1)\ \left\{ \begin{matrix} y_{1} = - 7 \\ x_{1} = 10 \\ \end{matrix} \right.\ \text{\ \ }\]
\[2)\left\{ \begin{matrix} y_{2} = 6\ \ \ \\ x_{2} = - 3 \\ \end{matrix} \right.\ .\]
\[\textbf{б)}\ \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \\ x + y^{2} = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + (1 + x)^{2} = - 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x + 1 + 2x + x^{2} + 1 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = 1 + x\ \ \ \ \ \ \ \ \ \\ x^{2} + 3x + 2 = 0 \\ \end{matrix} \right.\ \]
\[x^{2} + 3x + 2 = 0\]
\[x_{1} + x_{2} = - 3;\ \ \ x_{1} \cdot x_{2} = 2\]
\[x_{1} = - 1;\ \ \ x_{2} = - 2.\]
\[1)\ \left\{ \begin{matrix} x_{1} = - 1 \\ y_{1} = 0\ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]
\[2)\ \left\{ \begin{matrix} x_{2} = - 2 \\ y_{2} = - 1 \\ \end{matrix}. \right.\ \]
\[\textbf{в)}\ \left\{ \begin{matrix} x^{2} + y = 14 \\ y - x = 8\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x + 8\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + x + 8 - 14 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y = x + 8\ \ \ \ \ \ \ \ \ \\ x^{2} + x - 6 = 0 \\ \end{matrix} \right.\ \]
\[x^{2} + x - 6 = 0\]
\[x_{1} + x_{2} = - 1;\ \ \ x_{1} \cdot x_{2} = - 6\]
\[x_{1} = 2;\ \ \ \ x_{2} = - 3.\]
\[1)\ \left\{ \begin{matrix} x_{1} = 2\ \ \\ y_{1} = 10 \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x_{2} = - 3 \\ y_{2} = 5\ \ \ \\ \end{matrix} \right.\ .\]
\[\textbf{г)}\ \left\{ \begin{matrix} x + y = 4\ \ \\ y + xy = 6 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y + y(4 - y) = 6 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y + 4y - y^{2} - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 4 - y\ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 5y + 6 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} - 5y + 6 = 0\]
\[y_{1} + y_{2} = 5;\ \ \ y_{1} \cdot y_{2} = 6\]
\[y_{1} = 2;\ \ \ y_{2} = 3.\]
\[1)\ \left\{ \begin{matrix} y_{1} = 2 \\ x_{1} = 2 \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} y_{2} = 3 \\ x_{2} = 1. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (10;\ - 7);( - 3;6);\]
\[\textbf{б)}\ ( - 1;0);( - 2;\ - 1);\]
\[\textbf{в)}\ (2;10);\ \ ( - 3;5);\]
\[\textbf{г)}\ (2;2);(1;3).\]
\(\boxed{\text{430.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)
\[Пусть\ первому\ комбайнеру\ \]
\[требуется\ \text{x\ }часов,\ \]
\[а\ второму - \text{y\ }часов.\]
\[Обем\ работы\ равен\ 1.\]
\[Один\ справится\ на\ 24\ часа\ \]
\[быстрее,\ чем\ второй:\]
\[x + 24 = y.\]
\[\frac{1}{x} - производительность\ \]
\[первого\ комбайнера;\]
\[\frac{1}{y} - производительность\ \]
\[второго;\]
\[Вместе\ они\ выполнят\ \]
\[работу\ за\ 35\ ч:\]
\[35 \cdot \left( \frac{1}{x} + \frac{1}{y} \right) = 1.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} x + 24 = y\ \ \ \ \ \ \ \ \ \\ 35 \cdot \left( \frac{1}{x} + \frac{1}{y} \right) = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x + 24 = y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 35 \cdot \left( \frac{x + 24 + x}{x(x + 24)} \right) = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x + 24 = y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 35 \cdot (2x + 24) = x(x + 24) \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x + 24 = y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 70x + 840 = x^{2} + 24x \\ \end{matrix} \right.\ \]
\[x^{2} - 46x - 840 = 0\]
\[D_{1} = 23^{2} + 840 = 1369\]
\[x_{1,2} = 23 \pm 37 = 60;\ - 14.\]
\[Так\ как\ x > 0:\]
\[x = 60 \Longrightarrow y = 84.\]
\[Ответ:60\ ч\ и\ 84\ ч.\]
