Решебник по алгебре 9 класс Мерзляк Задание 462

\[\boxed{\text{462\ (462).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left\{ \begin{matrix} x + y - xy = 1 \\ \text{xy}(x + y) = 20 \\ \end{matrix} \right.\ \]

\[Пусть\ xy = a;\ \ x + y = b,\ тогда:\]

\[\left\{ \begin{matrix} b - a = 1 \\ ab = 20\ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} b = 1 + a\ \ \ \ \ \ \ \ \ \ \ \\ a + a^{2} - 20 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} b = - 4 \\ a = - 5 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} b = 5 \\ a = 4 \\ \end{matrix} \right.\ \]

\[a_{1} + a_{2} = - 1\]

\[a_{1}a_{2} = - 20\]

\[\left\{ \begin{matrix} a_{1} = - 5 \\ a_{2} = 4\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} xy = - 5\ \ \ \ \ \\ x + y = - 4 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \ \ \ \ \ \ \ \ \ \ \]

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

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

\[\ \left\{ \begin{matrix} x = 4 \\ y = 1 \\ \end{matrix} \right.\ \]

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

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

\[Ответ:(1; - 5),\ (1;4),\ \]

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

\[2)\ \left\{ \begin{matrix} \frac{y}{x} - \frac{x}{y} = \frac{21}{10} \\ x + y = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} \frac{y}{x} - \frac{x}{y} = \frac{21}{10} \\ x = 3 - y\ \\ \end{matrix} \right.\ \]

\[\frac{y}{3 - y} - \frac{3 - y}{y} - \frac{21}{10} = 0\]

\[21^{2} - 3y - 90 = 0\ \ \ |\ :3,\ \ \]

\[y \neq 3\]

\[7y^{2} - y - 30 = 0\]

\[D = 1 + 840 = 841\]

\[y = \frac{1 + 29}{14} = \frac{15}{7}\]

\[y = \frac{1 - 29}{14} = - 2\]

\[\left\{ \begin{matrix} x = \frac{6}{7}\text{\ \ } \\ y = \frac{15}{7} \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} x = 5\ \ \ \\ y = - 2 \\ \end{matrix} \right.\ \]

\[Ответ:\left( \frac{6}{7};\frac{15}{7} \right);\ (5;\ - 2).\]

\[3)\ \left\{ \begin{matrix} \frac{x}{y} + \frac{6y}{x} = 5\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 4xy - 3y^{2} = 18 \\ \end{matrix} \right.\ \]

\[Пусть\frac{x}{y} = a,\frac{y}{x} = \frac{1}{a}\]

\[a + \frac{6}{a} = 5\]

\[a^{2} + 6 - 5a = 0\]

\[a_{1} + a_{2} = 5\]

\[a_{1}a_{2} = 6\]

\[\left\{ \begin{matrix} a_{1} = 2 \\ a_{2} = 3 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{x}{y} = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 4xy - 3y^{2} - 18 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 4y^{2} + 8y^{2} - 3y^{2} - 18 = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 2y \\ 9y^{2} = 18 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 2y \\ y^{2} = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

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

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

\[\left\{ \begin{matrix} \frac{x}{y} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} + 4xy - 3y^{2} - 18 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} + 12y^{2} - 3y^{2} - 18 = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \\ 18y^{2} = 18 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 3y \\ y^{2} = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

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

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

\[Ответ:\left( 2\sqrt{2};\sqrt{2} \right),\ \left( - 2\sqrt{2};\ - \sqrt{2} \right),\ \]

\[(3;1),\ ( - 3;\ - 1).\]

\[4)\ \left\{ \begin{matrix} \frac{1}{x} + \frac{1}{y} = \frac{5}{6} \\ \frac{1}{x} - \frac{1}{y} = \frac{1}{6} \\ \end{matrix} \right.\ \]

\[Пусть\ \frac{1}{x} = a,а\ \ \ \frac{1}{y} = b,\ тогда:\ \]

\[\left\{ \begin{matrix} a + b = \frac{5}{6} \\ a - b = \frac{1}{6} \\ \end{matrix}\ ( - ) \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} 2b = \frac{2}{3}\text{\ \ \ \ \ \ } \\ a - b = \frac{1}{6} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ }\left\{ \begin{matrix} b = \frac{1}{3} \\ a = \frac{1}{2} \\ \end{matrix} \right.\ \]

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

\[Ответ:(2;3).\]

\[5)\ \left\{ \begin{matrix} \frac{y}{x} + xy = - 10 \\ \frac{5y}{x} - 2xy = 13 \\ \end{matrix} \right.\ \]

\[Пусть\ \frac{y}{x} = a,\ а\ xy = b,\ то:\]

\[\left\{ \begin{matrix} a + b = - 10\ \ \ \ | \cdot 2 \\ 5a - 2b = 13\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} 2a + 2b = - 20 \\ 5a - 2b = 13 \\ \end{matrix} + \right.\ \]

\[\left\{ \begin{matrix} 7a = - 7 \\ 5a - 2b = 13 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} a = - 1 \\ b = - 9 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} \frac{y}{x} = - 1 \\ xy = - 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} y = - x \\ - x^{2} = - 9 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

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

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

\[Ответ:(3;\ - 3),\ ( - 3;3).\]

\[6)\ \left\{ \begin{matrix} x^{2}y^{2} + xy = 6 \\ 2x - y = 3\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

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

\[Пусть\ xy = a,\ то:\]

\[a^{2} + a - 6 = 0\]

\[a_{1} + a_{2} = - 1,\ \ a_{1} = - 3\]

\[a_{1}a_{2} = - 6,\ \ a_{2} = 2\]

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

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

\[D = 9 - 24 < 0\]

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

\[2x^{2} - 3x - 2 = 0\]

\[D = 9 + 16 = 25\]

\[x = \frac{3 + 5}{4} = 2\]

\[x = \frac{3 - 5}{4} = - \frac{1}{2}\]

\[\left\{ \begin{matrix} x = 2 \\ y = 1 \\ \end{matrix} \right.\ \ \ \ или\ \ \ \left\{ \begin{matrix} x = - 0,5 \\ y = - 4\ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:(2;1);\ ( - 0,5;\ - 4).\]

\[7)\ \left\{ \begin{matrix} 3 \cdot (x + y)^{2} + 2 \cdot (x - 2y)^{2} = 5 \\ 2 \cdot (x - 2y) - x - y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \ \]

\[\left\{ \begin{matrix} 3 \cdot (x + y)^{2} + 2 \cdot (x - 2y)^{2} = 5 \\ 2 \cdot (x - 2y) - (x + y) = 1\ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\ \left\{ \begin{matrix} x + y = a \\ x - 2y = b \\ \end{matrix} \right.\ ,\ тогда:\]

\[\left\{ \begin{matrix} 3a^{2} + 2b^{2} = 5 \\ 2b - a = 1\ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} 3a^{2} + 2b^{2} = 5 \\ a = 2b - 1\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[3 \cdot (2b - 1)^{2} + 2b^{2} - 5 = 0\]

\[3 \cdot \left( 4b^{2} - 4b + 1 \right) + 2b^{2} - 5 = 0\]

\[12b^{2} - 12b + 3 + 2b^{2} - 5 = 0\]

\[14b^{2} - 12b - 2 = 0\ \ \ \ \ \ \ \ \ |\ :2\]

\[7b^{2} - 6b - 1 = 0\]

\[D = 36 + 28 = 64\]

\[b = \frac{6 + 8}{14} = 1\]

\[b = \frac{6 - 8}{14} = - \frac{1}{7}\]

\[\left\{ \begin{matrix} b = 1 \\ a = 1 \\ \end{matrix} \right.\ \ \ \ \ или\ \ \ \left\{ \begin{matrix} b = - \frac{1}{7} \\ a = - \frac{9}{7} \\ \end{matrix} \right.\ \]

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

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

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

\[\left\{ \begin{matrix} x + y = - \frac{9}{7} \\ x - 2y = - \frac{1}{7} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = - \frac{9}{7} - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - \frac{9}{7} - y - 2y + \frac{1}{7} = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\text{\ \ \ }\left\{ \begin{matrix} x = - \frac{9}{7} - y \\ y = - \frac{8}{21}\text{\ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - \frac{19}{21} \\ y = - \frac{8}{21} \\ \end{matrix} \right.\ \]

\[Ответ:(1;0);\ \left( - \frac{19}{21};\ - \frac{8}{21} \right)\text{.\ }\]