\[\boxed{\text{475\ (475).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[Пусть\ x - плотность\ первой\ \]
\[жидкости;а\ y - плотность\ \]
\[второй.\]
\[Плотность\ одной\ жидкости\ \]
\[на\ 0,2\ \frac{г}{см^{3}}\ больше\ другой:\]
\[x = y - 0,2.\]
\[12 + 14 = 26\ (г) - общая\ \]
\[масса\ двух\ жидкостей.\]
\[Составим\ систему\ уравнений:\]
\[\left\{ \begin{matrix} x = y - 0,2\ \ \ \\ \frac{12 + 14}{\frac{12}{x} + \frac{14}{y}} = 1,3 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = y - 0,2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 26xy = 1,3 \cdot (12y + 14x) \\ \end{matrix} \right.\ \Longrightarrow\]
\[26y^{2} - 39y + 3,64 = 0\]
\[y^{2} - 1,5y + 0,14 = 0\]
\[D = {1,5}^{2} - 4 \cdot 0,14 = 1,69\]
\[y_{1,2} = \frac{1,5 \pm 1,3}{2} = 1,4;0,1;\]
\[\left\{ \begin{matrix} y_{1} = 1,4 \\ x_{1} = 1,2 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \]
\[\left\{ \begin{matrix} y_{2} = 0,1\ \ \ \\ x_{2} = - 0,1 \\ \end{matrix} \right.\ \Longrightarrow не\ подходит,\ \]
\[так\ как\ x > 0.\]
\[Ответ:1,2\ \frac{г}{см^{3}}и\ 1,4\ \frac{г}{см^{3}}.\]
\[\boxed{\text{475.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \left\{ \begin{matrix} x^{2} + 3xy - 10y^{2} = 0 \\ x^{2} - 4xy + 3y = 0\ \ \ \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x + 5y)(x - 2y) = 0 \\ x² - 4xy + 3y = 0\ \ \ \ \ \ \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x + 5y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4xy + 3y = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 25y^{2} + 20y^{2} + 3y = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 5y\ \ \ \ \ \ \ \ \ \ \\ 45y^{2} + 3y = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 5y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 3y(15y + 1) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 0 \\ x_{1} = 0 \\ \end{matrix} \right.\ \ \ или\left\{ \begin{matrix} y_{2} = - \frac{1}{15} \\ x_{2} = \frac{1}{3}\text{.\ \ \ \ \ } \\ \end{matrix} \right.\ \]
\[2)\ \left\{ \begin{matrix} x - 2y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 4xy + 3y = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - 8y^{2} + 3y = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ y(3 - 4y) = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = 0 \\ x_{1} = 0 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \left\{ \begin{matrix} y_{2} = 0,75 \\ x_{2} = 1,5.\ \ \\ \end{matrix} \right.\ \]
\[\textbf{б)}\ \left\{ \begin{matrix} x^{2} + xy - 6y^{2} = 0\ \ \ \ \ \ \ \\ x^{2} + 3xy + 2y - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} (x + 3y)(x - 2y) = 0\ \ \\ x² + 3xy + 2y - 6 = 0 \\ \end{matrix} \right.\ \]
\[1)\ \left\{ \begin{matrix} x + 3y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3xy + 2y - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = - 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} - 9y^{2} + 2y - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \ \left\{ \begin{matrix} x = - 3y \\ 2y = 6\ \ \\ \end{matrix} \right.\ \Longrightarrow \left\{ \begin{matrix} y = 3\ \ \\ x = - 9 \\ \end{matrix} \right.\ ;\]
\[2)\ \left\{ \begin{matrix} x - 2y = 0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 3xy + 2y - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} + 6y^{2} + 2y - 6 = 0 \\ \end{matrix} \right.\ \Longrightarrow\]
\[\Longrightarrow \left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5y² + y - 3 = 0 \\ \end{matrix} \right.\ \]
\[5y^{2} + y - 3 = 0\]
\[D = 1 + 4 \cdot 5 \cdot 3 = 61\]
\[y_{1,2} = \frac{- 1 \pm \sqrt{61}}{10};\ \]
\[\Longrightarrow \left\{ \begin{matrix} y_{1} = \frac{- 1 + \sqrt{61}}{5} \\ x_{1} = \frac{- 1 + \sqrt{61}}{10} \\ \end{matrix} \right.\ \ \ \ \ или\]
\[\left\{ \begin{matrix} y_{2} = \frac{- 1 - \sqrt{61}}{5} \\ x_{2} = \frac{- 1 - \sqrt{61}}{10}. \\ \end{matrix} \right.\ \]
\[Ответ:а)\ (0;0);(1,5;0,75);\]
\[\left( \frac{1}{3};\ - \frac{1}{15} \right);\]
\[\ \ \ \ \ \ \ \ \ \ \ б)\ ( - 9;3);\]
\[\left( \frac{- 1 + \sqrt{61}}{5};\frac{- 1 + \sqrt{61}}{10} \right);\]
\[\left( \frac{- 1 - \sqrt{61}}{5};\frac{- 1 - \sqrt{61}}{10} \right)\text{.\ }\]