Решебник по алгебре 9 класс Мерзляк Задание 815

\[\boxed{\mathbf{815\ (815).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[600\ \frac{м}{мин} = \frac{600 \cdot 60}{1000} =\]

\[= 36\ \frac{км}{ч}.\]

\[Пусть\ x\frac{км}{ч} - скорость\ \]

\[велосипедиста,\ тогда\ \]

\[(x + 36)\ \frac{км}{ч} - скорость\]

\[мотоциклиста.\ Значит,\]

\[\ \left( \frac{120}{x} \right)\ ч - едет\ велосипедист,\ \]

\[\left( \frac{120}{x + 36} \right)\ ч - едет\]

\[мотоциклист.\]

\[Составляем\ уравнение:\]

\[\frac{120}{x} - \frac{120}{x + 36} = 3\]

\[\frac{120}{x} - \frac{120}{x + 36} - 3 = 0\ \ \ |\ :3\]

\[\frac{40}{x} - \frac{40}{x + 36} - 1 = 0\ \ \]

\[\ | \cdot (x(x + 36)) \neq 0\]

\[40 \cdot (x + 36) - 40x -\]

\[- x(x + 36) = 0\]

\[40x - 1440 - 40x - x^{2} -\]

\[- 36x = 0\ \ \ \ | \cdot ( - 1)\]

\[x^{2} + 36x - 1440 = 0\]

\[x_{1} + x_{2} = - 36,\ \ \]

\[x_{1} = - 60 \Longrightarrow не\ удовлетворяет.\]

\[x_{1}x_{2} = - 1440,\ \ \]

\[x_{2} = 24\ \left( \frac{км}{ч} \right) - скорость\]

\[\ велосипедиста.\]

\[Ответ:24\ \frac{км}{ч}.\]

\[\boxed{\mathbf{815.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\left\{ \begin{matrix} x^{2} + y^{2} = 5 \\ xy = 2\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} \left( \frac{2}{y} \right)^{2} + y^{2} = 5 \\ x = \frac{2}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} \frac{4}{y^{2}} + y^{2} = 5 \\ x = \frac{2}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ | \cdot y^{2} \neq 0\]

\[4 + y^{4} - 5y^{2} = 0\]

\[\left( y^{2} - 4 \right)\left( y^{2} - 1 \right) = 0\]

\[(y - 2)(y + 2)(y - 1)(y + 1) =\]

\[\left\{ \begin{matrix} y = 2 \\ x = 1 \\ \end{matrix}\text{\ \ \ \ } \right.\ \ \left\{ \begin{matrix} y = - 2 \\ x = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} y = 1 \\ x = 2 \\ \end{matrix} \right.\ \ \]

\[\text{\ \ \ }\left\{ \begin{matrix} y = - 1 \\ x = - 2 \\ \end{matrix} \right.\ \]

\[Ответ:(1;2),\ ( - 1;\ - 2),\ (2;1),\]

\[\ ( - 2;\ - 1).\]

\[2)\ \left\{ \begin{matrix} xy + x + y = 11 \\ \text{xy}(x + y) = 30\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x + y = 11 - xy \\ \text{xy}(11 - xy) = 30 \\ \end{matrix} \right.\ \]

\[11xy - \left( \text{xy} \right)^{2} - 30 = 0\]

\[\left( \text{xy} \right)^{2} - 11xy + 30 = 0\]

\[xy = 5,\ \ xy = 6\]

\[\left\{ \begin{matrix} xy = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 5 \cdot (x + y) = 30 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} xy = 5\ \ \ \ \ \ \\ x + y = 6 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} (6 - y)y = 5 \\ x = 6 - y\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ }\]

\[\ \left\{ \begin{matrix} y^{2} - 6y + 5 = 0 \\ x = 6 - y\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 1 \\ x = 5 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \ \left\{ \begin{matrix} y = 5 \\ x = 1 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} xy = 6\ \ \ \ \ \\ x + y = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} (5 - y)y = 6 \\ x = 5 - y\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y^{2} - 5y + 6 = 0 \\ x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y = 3 \\ x = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ }или\text{\ \ }\left\{ \begin{matrix} y = 2 \\ x = 3 \\ \end{matrix} \right.\ \]

\[Ответ:(5;1),\ (1;5),\ (2;3),\ (3;2).\]

\[3)\ \left\{ \begin{matrix} xy + x + y = 5 \\ x - xy + y = 1 \\ \end{matrix} \right.\ + \ \ \ \]

\[\left\{ \begin{matrix} 2x + 2y = 6\ \ \ |\ :2 \\ x + y + xy = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} x + y = 3 \\ 3 + xy = 5 \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} x + y = 3 \\ xy = 2\ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \\ y(3 - y) = 2 \\ \end{matrix} \right.\ \]

\[y^{2} - 3y + 2 = 0\]

\[y_{1} + y_{2} = 3,\ \ y_{1} = 2\]

\[y_{1}y_{2} = 2,\ \ y_{2} = 1\]

\[\left\{ \begin{matrix} y = 2 \\ x = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \ \ \ \ \left\{ \begin{matrix} y = 1 \\ x = 2 \\ \end{matrix} \right.\ \]

\[Ответ:(1;2),\ (2;1).\]

\[4)\ \left\{ \begin{matrix} x - y = 2 \\ \frac{x}{y} - \frac{y}{x} = \frac{5}{6} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x - y = 2 \\ \frac{x^{2} - y^{2}}{\text{xy}} = \frac{5}{6} \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x - y = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{(x - y)(x + y)}{\text{xy}} = \frac{5}{6} \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x - y = 2\ \ \ \ \ \ \ \ \ \\ \frac{2 \cdot (x + y)}{\text{xy}} = \frac{5}{6} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 12 \cdot (x + y) = 5xy \\ x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 12 \cdot (2 + y + y) = 5y(2 + y) \\ x = 2 + y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[24 + 24y = 10y + 5y^{2}\]

\[5y^{2} - 14y - 24 = 0\]

\[D = 196 + 480 = 676\]

\[y = \frac{14 + 26}{10} = 4,\]

\[\ \ y = \frac{14 - 26}{10} = - 1,2\]

\[\left\{ \begin{matrix} y = 4 \\ x = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \ \ \ \ \left\{ \begin{matrix} y = - 1,2 \\ x = 0,8\ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:(6;4),\ (0,8;\ - 1,2).\]

\[5)\ \left\{ \begin{matrix} x^{2} + 2xy + y^{2} = 25 \\ x^{2} - 3xy = 4\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} (x + y)^{2} = 25 \\ x^{2} - 3xy = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} x + y = 5\ \ \ \ \ \ \\ x^{2} - 3xy = 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ (5 - y)^{2} - 3y(5 - y) = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 25 - 10y + y^{2} - 15y + 3y^{2} = 4 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} - 25y + 21 = 0 \\ \end{matrix} \right.\ ,\]

\[\ \ x = 5 - y\]

\[D = 625 - 336 = 289\]

\[y = \frac{25 + 17}{8} = \frac{21}{4},\ \ \]

\[y = \frac{25 - 17}{8} = 1\]

\[\left\{ \begin{matrix} y = 1 \\ x = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \ \ \left\{ \begin{matrix} y = 5,25 \\ x = 0,25 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x + y = - 5 \\ x^{2} - 3xy = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} x = - 5 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ ( - 5 - y)^{2} - 3y( - 5 - y) = 4 \\ \end{matrix} \right.\ \]

\[25 + 10y + y^{2} + 15y +\]

\[+ 3y^{2} - 4 = 0\]

\[4y^{2} + 25y + 21 = 0\]

\[D = 625 - 4 \cdot 4 \cdot 21 = 289\ \ \]

\[y = \frac{- 25 + 17}{8} = - 1;\ \ \]

\[y = \frac{- 25 - 17}{8} = - 5,25\]

\[\left\{ \begin{matrix} y = - 1 \\ x = - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ }или\ \ \ \ \ \left\{ \begin{matrix} y = - 5,25 \\ x = 0,25 \\ \end{matrix} \right.\ \]

\[Ответ:(4;1),\ (0,25;5,25),\ \]

\[( - 4;\ - 1),\ (0,25;\ - 5,25).\]

\[6)\ \left\{ \begin{matrix} 2x^{2} - y^{2} = 14 \\ xy = - 6\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2x^{2} - \left( - \frac{6}{x} \right)^{2} = 14 \\ y = - \frac{6}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} 2x² - \frac{36}{x^{2}} - 14 = 0\ \ | \cdot x^{2} \\ y = - \frac{6}{x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[2x^{4} - 14x^{2} - 36 = 0\ \ \ |\ :2\]

\[x^{4} - 7x^{2} - 18 = 0\]

\[x^{2} = 9,\]

\[\ \ x² = - 2\ \ (не\ удовлетворяет).\]

\[\left\{ \begin{matrix} x^{2} = 9 \\ y = - \frac{6}{x}\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 3 \\ y = - 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \]

\[\text{\ \ \ \ }\left\{ \begin{matrix} x = - 3 \\ y = 2\ \\ \end{matrix} \right.\ \]

\[Ответ:(3;\ - 2),\ ( - 3;2).\]

\[7)\ \left\{ \begin{matrix} \frac{1}{x} - \frac{1}{y} = \frac{1}{10} \\ xy = 50\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{y - x}{\text{xy}} = \frac{1}{10}\ \\ xy = 50\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} \frac{y - x}{50} = \frac{1}{10} \\ xy = 50\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 10 \cdot (y - x) = 50\ \ \ |\ :10 \\ x = \frac{50}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} y - x = 5 \\ x = \frac{50}{y}\text{\ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y - \frac{50}{y} = 5\ \ \ | \cdot y \neq 0 \\ x = \frac{50}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[y² - 5y - 50 = 0\]

\[y_{1} + y_{2} = 5,\ \ y_{1} = 10\]

\[y_{1}y_{2} = - 50,\ \ y_{2} = - 5\]

\[\left\{ \begin{matrix} x = 5 \\ y = 10 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \ \ \ \ \ \left\{ \begin{matrix} x = - 10 \\ y = - 5 \\ \end{matrix} \right.\ \]

\[Ответ:(5;10),\ ( - 10;\ - 5).\]

\[8)\ \left\{ \begin{matrix} \frac{x}{y} - \frac{y}{x} = \frac{16}{15} \\ x^{2} - y^{2} = 16 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{2} - y^{2}}{\text{xy}} = \frac{16}{15} \\ x^{2} - y^{2} = 16 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{16}{\text{xy}} = \frac{16}{15}\ \ \ \ \ |\ :16 \\ x^{2} - y^{2} = 16\ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{1}{\text{xy}} = \frac{1}{15}\text{\ \ \ \ \ \ \ \ \ } \\ x^{2} - y^{2} = 16 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} xy = 15\ \ \ \ \ \ \ \ \\ x^{2} - y^{2} = 16 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} x = \frac{15}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{225}{y^{2}} - y^{2} - 16 = 0\ \ | \cdot ( - y^{2} \neq 0) \\ \end{matrix} \right.\ \]

\[y^{4} + 16y^{2} - 225 = 0\]

\[y^{2} = 9,\]

\[\ \ y² = - 25\ (не\ удовлетворяет)\]

\[\left\{ \begin{matrix} y^{2} = 9 \\ x = \frac{15}{y} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 3 \\ x = 5 \\ \end{matrix} \right.\ \text{\ \ \ }или\ \ \ \]

\[\text{\ \ }\left\{ \begin{matrix} y = - 3 \\ x = - 5 \\ \end{matrix} \right.\ \]

\[Ответ:(5;3),\ ( - 5;\ - 3).\]

\[9)\ \left\{ \begin{matrix} \frac{x}{y} + \frac{y}{x} = \frac{25}{12}\text{\ \ \ \ } \\ x^{2} + y^{2} = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} \frac{x^{2} + y^{2}}{\text{xy}} = \frac{25}{12} \\ x^{2} + y^{2} = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} \frac{25}{\text{xy}} = \frac{25}{12}\ \ \ \ \ |\ :25 \\ x² + y² = 25\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{1}{\text{xy}} = \frac{1}{12}\text{\ \ \ \ \ \ \ \ \ } \\ x^{2} + y^{2} = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} xy = 12\ \ \ \ \ \ \ \ \ \\ x^{2} + y^{2} = 25 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{12}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \left( \frac{12}{y} \right)^{2} + y^{2} - 25 = 0 \\ \end{matrix} \right.\ \text{\ \ }\]

\[\left\{ \begin{matrix} x = \frac{12}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{144}{y^{2}} + y² - 25 = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \frac{12}{y}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ y^{4} - 25y^{2} + 144 = 0 \\ \end{matrix} \right.\ ;\ \ \]

\[y^{2} = 16,\ \ y^{2} = 9\]

\[\left\{ \begin{matrix} y^{2} = 16 \\ x = \frac{12}{y}\text{\ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 4 \\ x = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} y = - 4 \\ x = - 3 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y^{2} = 9 \\ x = \frac{12}{y} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} y = 3 \\ x = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} y = - 3 \\ x = - 4 \\ \end{matrix} \right.\ \]

\[Ответ:(3;4),\ ( - 3;\ - 4),\ (4;3),\ \]

\[( - 4;\ - 3).\]

\[10)\ \left\{ \begin{matrix} x^{2} + 2xy = 5 \\ y^{2} - 4xy = - 4 \\ \end{matrix} \right.\ \ \ + ,\]

\[\text{\ \ }x^{2} - 2xy + y^{2} = 1,\ \ \]

\[(x - y)^{2} = 1\]

\[\left\{ \begin{matrix} x - y = 1\ \ \ \ \ \ \ \ \ \\ y^{2} - 4xy = - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 4y(y + 1) = - 4 \\ \end{matrix} \right.\ \]

\[\text{\ \ \ \ }\left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 4y^{2} - 4y + 4 = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = y + 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 3y^{2} - 4y + 4 = 0\ \ \ | \cdot ( - 1) \\ \end{matrix} \right.\ \]

\[3y^{2} + 4y - 4 = 0\]

\[D = 16 + 48 = 64\]

\[y = \frac{- 4 + 8}{6} = \frac{2}{3},\ \ \]

\[y = \frac{- 4 - 8}{6} = - 2\]

\[\left\{ \begin{matrix} x = 1\frac{2}{3} \\ y = \frac{2}{3}\text{\ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }или\ \ \ \ \left\{ \begin{matrix} x = - 1 \\ y = - 2 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 4y(y - 1) = - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} - 4y^{2} + 4y + 4 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\left\{ \begin{matrix} x = y - 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 3y^{2} + 4y + 4 = 0\ \ | \cdot ( - 1) \\ \end{matrix} \right.\ \]

\[3y^{2} - 4y - 4 = 0\]

\[D = 16 + 48 = 64\]

\[y = \frac{4 + 8}{6} = 2,\]

\[\ \ y = \frac{4 - 8}{6} = - \frac{2}{3}\]

\[\left\{ \begin{matrix} x = 1 \\ y = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }или\ \ \ \ \left\{ \begin{matrix} x = - 1\frac{2}{3} \\ y = - \frac{2}{3}\text{\ \ } \\ \end{matrix} \right.\ \]

\[Ответ:(1;2),\ \left( - 1\frac{2}{3};\ - \frac{2}{3} \right),\ \]

\[\left( 1\frac{2}{3};\frac{2}{3} \right),\ ( - 1;\ - 2).\]