\[\boxed{\text{431\ (431).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ \left\{ \begin{matrix} x - y = 3 \\ xy = 2\ \ \ \ \ \ \\ \end{matrix} \Longrightarrow \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 3\ \ \ \ \ \ \ \ \ \\ y(y + 3) = - 2 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 3\ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 3y + 2 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} + 3y + 2 = 0\]
\[y_{1} + y_{2} = - 3;\ \ \ y_{1} \cdot y_{2} = 2\]
\[y_{1} = - 1;\ \ y_{2} = - 2;\]
\[1)\ \left\{ \begin{matrix} y_{1} = - 1 \\ x_{1} = 2\ \ \\ \end{matrix} \right.\ \text{\ \ }\]
\[2)\ \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = 1\ \ \\ \end{matrix}. \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x + y = 2,5 \\ xy = 1,5\ \ \ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2,5 - y\ \ \ \ \ \ \ \ \ \\ y(2,5 - y) = 1,5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2,5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2,5y - y^{2} - 1,5 = 0 \\ \end{matrix} \right.\ \]
\[y^{2} - 2,5y + 1,5 = 0\ \ \ \ \ \]
\[y_{1} + y_{2} = 2,5;\ \ \ \ y_{1} \cdot y_{2} = 1,5\]
\[y_{1} = 1;\ \ \ \ \ \ \ y_{2} = 1,5;\]
\[1)\ \left\{ \begin{matrix} y_{1} = 1\ \ \ \ \\ x_{1} = 1,5 \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} y_{2} = 1,5 \\ x_{2} = 1\ \ \ \ \\ \end{matrix}. \right.\ \]
\[\textbf{в)}\ \left\{ \begin{matrix} x + y = - 1 \\ x^{2} + y^{2} = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + ( - 1 - y)^{2} = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 1 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 1 + 2y + y^{2} = 1 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 1 - y\ \ \\ 2y^{2} + 2y = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 1 - y \\ y(y + 1) = 0 \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} y_{1} = 0\ \ \ \\ x_{1} = - 1 \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} y_{2} = - 1 \\ x_{2} = 0\ \ \ \\ \end{matrix} \right.\ .\]
\[\textbf{г)}\ \left\{ \begin{matrix} x - y = 2\ \ \ \ \ \ \ \\ x^{2} - y^{2} = 17 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (y + 2)^{2} - y^{2} = 17 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + 4y + 4 - y^{2} - 17 = 0 \\ \end{matrix} \right.\ \]
\[\Longrightarrow \left\{ \begin{matrix} x = y + 2 \\ 4y = 13\ \ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = 3\frac{1}{4} \\ x = 5\frac{1}{4}. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (2;\ - 1);(1;\ - 2);\ \ \]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ б)\ (1,5;1);(1;1,5);\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ в)\ ( - 1;0);(0;\ - 1);\]
\[\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ г)\ \left( 5\frac{1}{4};3\frac{1}{4} \right).\]
\(\boxed{\text{431.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\)
\[Пусть\ первая\ бригада\ затратит\ \]
\[на\ работу\ \text{x\ }(ч),\ а\ вторая - \text{y\ }ч.\ \]
\[Одна\ бригада\ заасфальтирует\ \]
\[на\ 4\ ч\ быстрее,\ чем\ другая:\]
\[x - 4 = y.\]
\[\frac{1}{x} - производительность\ \]
\[первой\ бригады;\]
\[\frac{1}{y} - производительность\ \]
\[второй\ бригады;\]
\[Работая\ вместе\ 24\ ч\ обе\ \]
\[бригады\ заасфальтировали\ \]
\[5\ участков:\]
\[24 \cdot \left( \frac{1}{x} + \frac{1}{y} \right) = 5.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} x - 4 = y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 24 \cdot \left( \frac{1}{x} + \frac{1}{y} \right) = 5 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x - 4 = y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 24 \cdot (x - 4 + x) = 5x(x - 4) \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x - 4 = y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5x^{2} - 68x + 96 = 0 \\ \end{matrix} \right.\ \]
\[5x^{2} - 68x + 96 = 0\]
\[D = 34^{2} - 5 \cdot 96 = 676\]
\[x_{1,2} = \frac{34 \pm 26}{5} = 12;1,6.\]
\[\left\{ \begin{matrix} x_{1} = 12 \\ y_{1} = 8\ \ \ \\ \end{matrix} \right.\ \ \ \ \ \ \ или\ \ \]
\[\left\{ \begin{matrix} x_{2} = 1,6\ \ \ \\ y_{2} = - 2,4 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow не\ подходит,\ так\ как\ y > 0.\]
\[Ответ:8\ ч\ и\ 12\ ч.\]