Решебник по алгебре 9 класс Мерзляк Задание 463

\[\boxed{\text{463\ (463).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ \left\{ \begin{matrix} \frac{x^{\backslash x}}{y} + \frac{y^{\backslash y}}{x} = 2,5 \\ 2x - 3y = 3\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

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

\[\left\{ \begin{matrix} x^{2} + y^{2} - 2,5xy = 0\ \ \ \ \ \ \ | \cdot 4 \\ (2x - 3y)^{2} = 9\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} 4x^{2} + 4y^{2} - 10xy = 0 \\ 4x^{2} - 12xy + 9y^{2} - 9 = 0 \\ \end{matrix} - \right.\ \]

\[\left\{ \begin{matrix} - 5y^{2} + 2xy + 9 = 0 \\ 2x - 3y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix}\ \right.\ \text{\ \ \ }\ \]

\[\left\{ \begin{matrix} - 5y^{2} + 3y + 3y^{2} + 9 = 0 \\ 2x = 3 + 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

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

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

\[- 2y^{2} + 3y + 9 = 0\]

\[D = 9 + 72 = 81\]

\[y_{1} = \frac{- 3 + 9}{- 4} = - \frac{3}{2}\]

\[y_{2} = \frac{- 3 - 9}{- 4} = 3\]

\[\left\{ \begin{matrix} y = 3 \\ x = 6 \\ \end{matrix} \right.\ \ \ \ \ \ \ \ \ \ или\ \ \ \ \ \ \ \ \left\{ \begin{matrix} y = - \frac{3}{2} \\ x = - \frac{3}{4} \\ \end{matrix} \right.\ \]

\[Ответ:(6;3),\ ( - 0,75;\ - 1,5).\]

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

\[\frac{x - 2y}{x + y} = a:\]

\[\frac{x + y}{x - 2y} = \frac{1}{a}\]

\[a - \frac{1}{4} = \frac{15}{4}\]

\[\frac{4a^{2} - 4 - 15a}{4a} = 0,\ \ a \neq 0\]

\[4a^{2} - 4 - 15a = 0\]

\[D = 225 + 64 = 289\]

\[a_{1} = \frac{15 + 17}{8} = 4\]

\[a_{2} = \frac{15 - 17}{8} = - \frac{1}{4}\]

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

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

\[\left\{ \begin{matrix} x - 4x - 2y - 4y = 0 \\ 4x + 5y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} - 3x - 6y = 0 \\ 4x + 5y = 3\ \ \ \\ \end{matrix} + \right.\ \text{\ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} - 3x + 4x - 6y + 5y = 3 \\ 4x + 5y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

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

\[\left\{ \begin{matrix} x = 3 + \text{y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 12 + 4y + 5y = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

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

\[\left\{ \begin{matrix} \frac{x - 2y}{x + y} = - \frac{1}{4} \\ 4x + 5y = 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }\]

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

\[\left\{ \begin{matrix} 4x - 8y + x + y = 0 \\ 4x + 5y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} 5x - 7y = 0 \\ 4x + 5y = 3 \\ \end{matrix} - \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 5x - 4x - 7y - 5y = - 3 \\ 4x + 5y = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x - 12y = - 3 \\ 4x + 5y = 3\ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = - 3 + 12y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ - 12 + 48y + 5y - 3 = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\text{\ \ }\left\{ \begin{matrix} x = - 3 + 12y \\ 53y = 15\ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = - 3 + 12y \\ y = \frac{15}{53}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = \frac{21}{53} \\ y = \frac{15}{53} \\ \end{matrix} \right.\ \]

\[Ответ:(2;\ - 1);\ \left( \frac{21}{53};\frac{15}{53} \right).\]

\[3)\ \left\{ \begin{matrix} \frac{1}{x} + \frac{4}{y} = 4 \\ \frac{1}{y} - \frac{2}{x} = 10 \\ \end{matrix} \right.\ \]

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

\[\left\{ \begin{matrix} a + 4b = 4\ \ \ | \cdot 2 \\ b - 2a = 10\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 2a + 8b = 8 \\ - 2a + b = 10 \\ \end{matrix} + \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[9b = 18,\ \ b = 2\]

\[\left\{ \begin{matrix} b = 2\ \ \ \\ a = - 4 \\ \end{matrix} \right.\ \]

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

\[Ответ:\left( - \frac{1}{4};\frac{1}{2} \right).\]

\[4)\ \left\{ \begin{matrix} \frac{x}{y} + \frac{y}{x} = \frac{10}{3}\text{\ \ \ \ } \\ x^{2} - y^{2} = 72 \\ \end{matrix} \right.\ \]

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

\[a + \frac{1}{a} = \frac{10}{3}\]

\[3a^{2} + 3 - 10a = 0,\ \ a \neq 0\]

\[D = 100 - 36 = 64\]

\[a = \frac{10 + 8}{6} = 3\]

\[a = \frac{10 - 8}{6} = \frac{1}{3}\]

\[\left\{ \begin{matrix} \frac{x}{y} = 3\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - y^{2} = 72 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = 3y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{2} - y^{2} = 72 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = 3y\ \ \ \ \ \\ 8y^{2} = 72 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 3y \\ y^{2} = 9 \\ \end{matrix} \right.\ \]

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

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

\[\left\{ \begin{matrix} \frac{x}{y} = \frac{1}{3}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ x^{2} - y^{2} = 72 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} y = 3x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{2} - 9x^{2} = 72 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 3x\ \ \ \ \ \ \ \\ - 8x^{2} = 72 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\]

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

\[решений\ \]

\[Ответ:(9;3);\ ( - 9;\ - 3).\]

\[5)\ \left\{ \begin{matrix} 4 \cdot (x - y)^{2} + 7 \cdot (x - y) = 15 \\ 2x + 5y = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\ x - y = a,\]

\[4a^{2} + 7a = 15\]

\[4a^{2} + 7a - 15 = 0\]

\[D = 49 + 240 = 289\]

\[a = \frac{- 7 + 17}{8} = \frac{10}{8} = \frac{5}{4}\]

\[a = \frac{- 7 - 17}{8} = - 3\]

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

\[\left\{ \begin{matrix} x = - 3 + y\ \ \ \ \ \ \ \ \ \ \ \ \\ - 6 + 2y + 5y = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} x = - 3 + y \\ 7y = 7\ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = - 2 \\ y = 1\ \ \ \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x - y = \frac{5}{4}\text{\ \ \ \ \ } \\ 2x + 5y = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{5}{4} + y\ \ \ \ \ \ \ \ \ \ \ \ \ \\ \frac{5}{2} + 2y + 5y = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x = \frac{5}{4} + y \\ 7y = - \frac{3}{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = \frac{5}{4} + y \\ y = - \frac{3}{14} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x = \frac{29}{28}\text{\ \ \ \ } \\ y = - \frac{3}{14} \\ \end{matrix} \right.\ \]

\[Ответ:\left( \frac{29}{28};\ - \frac{3}{14} \right);\ ( - 2;1).\]

\[6)\ \left\{ \begin{matrix} (x - y)^{2} + 2x = 35 + 2y \\ (x + y)^{2} + 2y = 3 - 2x\ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ \ }\]

\[\ \left\{ \begin{matrix} (x - y)^{2} + 2x - 35 - 2y = 0 \\ (x + y)^{2} + 2y - 3 + 2x = 0 \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} (x - y)^{2} + 2 \cdot (x - y) - 35 = 0 \\ (x + y)^{2} + 2 \cdot (x + y) - 3 = 0\ \ \\ \end{matrix} \right.\ \]

\[x - y = a\]

\[x + y = b\]

\[\left\{ \begin{matrix} a^{2} + 2a - 35 = 0 \\ b^{2} + 2b - 3 = 0 \\ \end{matrix} \right.\ \]