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МатематикаГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 267
Задание 267
\[\boxed{\mathbf{267}.}\]
\[1)\ 2x^{2} + 5xy - 12y^{2} = 6\]
\[2x^{2} - 3xy + 8xy - 12y^{2} = 6\]
\[x(2x - 3y) + 4y(2x - 3y) = 6\]
\[(x + 4y)(2x - 3y) = 6\]
\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ \pm 6.\]
\[\left\{ \begin{matrix} x + 4y = 1\ \ | \cdot 2 \\ 2x - 3y = 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 2x + 8y = 2 \\ 2x - 3y = 6 \\ \end{matrix} \right.\ ( - )\]
\[11y = - 4\]
\[y = - 4\ \ \Longrightarrow y \notin N.\]
\[\left\{ \begin{matrix} x + 4y = 6\ \ | \cdot 2 \\ 2x - 3y = 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 2x + 8y = 12 \\ 2x - 3y = 1\ \ \\ \end{matrix} \right.\ ( - )\]
\[11y = 11\]
\[y = 1.\]
\[x = 6 - 4y = 6 - 4 = 2.\]
\[\Longrightarrow (2;1).\]
\[\left\{ \begin{matrix} x + 4y = - 1\ \ | \cdot 2 \\ 2x - 3y = - 6\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]
\[\ \left\{ \begin{matrix} 2x + 8y = - 2 \\ 2x - 3y = - 6 \\ \end{matrix} \right.\ \ ( - )\]
\[11y = 4\]
\[y = \frac{4}{11}\]
\[\Longrightarrow y \notin N.\]
\[\left\{ \begin{matrix} x + 4y = - 6\ \ | \cdot 2 \\ 2x - 3y = - 1\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 2x + 8y = - 12 \\ 2x - 3y = - 1\ \ \\ \end{matrix} \right.\ ( - )\]
\[11y = - 11\]
\[y = - 1\]
\[\Longrightarrow y \notin N.\]
\[\left\{ \begin{matrix} x + 4y = 2\ \ | \cdot 2 \\ 2x - 3y = 3\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 2x + 8y = 4 \\ 2x - 3y = 3 \\ \end{matrix} \right.\ ( - )\]
\[11y = 1\]
\[y = \frac{1}{11}\]
\[\Longrightarrow y \notin N.\]
\[\left\{ \begin{matrix} x + 4y = - 2\ \ | \cdot 2 \\ 2x - 3y = - 3\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\left\{ \begin{matrix} 2x + 8y = - 4 \\ 2x - 3y = - 3 \\ \end{matrix} \right.\ \ ( - )\]
\[11y = - 1\]
\[y = - \frac{1}{11}\]
\[\Longrightarrow y \notin N.\]
\[\left\{ \begin{matrix} x + 4y = 3\ \ | \cdot 2 \\ 2x - 3y = 2\ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} 2x + 8y = 6 \\ 2x - 3y = 2 \\ \end{matrix} \right.\ ( - )\]
\[11y = 4\]
\[y = \frac{4}{11}\]
\[\Longrightarrow y \notin N.\]
\[\left\{ \begin{matrix} x + 4y = - 3\ \ | \cdot 2 \\ 2x - 3y = - 2\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} 2x + 8y = - 6 \\ 2x - 3y = - 2 \\ \end{matrix} \right.\ ( - )\]
\[11y = - 4\]
\[y = - \frac{4}{11}\]
\[\Longrightarrow y \notin N.\]
\[Ответ:(2;1).\]
\[2)\ 2x^{2} - xy - y^{2} +\]
\[+ 2x + 7y = 28\]
\[2x^{2} - 2xy + xy - y^{2} + 6x -\]
\[- 4x + 4y + 3y = 28\]
\[2x^{2} - 2xy + xy - y^{2} + 6x +\]
\[+ 3y - 4x + 4y - 12 + 12 = 28\]
\[2x(x - y + 3) +\]
\[+ y(x - y + 3) -\]
\[- 4(x - y + 3) = 16\]
\[(2x + y - 4)(x - y + 3) = 16\]
\[Делители:\ \pm 1; \pm 2;\ \pm 4;\ \pm 8;\ \]
\[\pm 16.\]
\[\left\{ \begin{matrix} x - y + 3 = 1\ \ \ \ \\ 2x + y - 4 = 16 \\ \end{matrix} \right.\ \text{\ \ \ }\left\{ \begin{matrix} x = 6 \\ y = 8 \\ \end{matrix} \right.\ ;\]
\[\left\{ \begin{matrix} x - y + 3 = 16 \\ 2x + y - 4 = 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = 6\ \ \ \\ y = - 7 \\ \end{matrix} \right.\ ;\]
\[\left\{ \begin{matrix} x - y + 3 = - 1\ \ \ \ \ \\ 2x + y - 4 = - 16 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = - \frac{16}{3} \\ y = - \frac{4}{3}\text{\ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - y + 3 = - 16 \\ 2x + y - 4 = - 1 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x = - \frac{16}{3} \\ y = \frac{41}{3}\text{\ \ \ \ } \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - y + 3 = 2\ \ \\ 2x + y - 4 = 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = \frac{11}{3} \\ y = \frac{14}{3} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - y + 3 = 8\ \ \\ 2x + y - 4 = 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ \ \ }\left\{ \begin{matrix} x = \frac{11}{3}\text{\ \ } \\ y = - \frac{4}{3} \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - y + 3 = - 2\ \ \ \ \\ 2x + y - 4 = - 8 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = - 3 \\ y = 2\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - y + 3 = - 8\ \ \\ 2x + y - 4 = - 2 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\left\{ \begin{matrix} x = - 3 \\ y = 8\ \ \ \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - y + 3 = 4\ \ \ \ \\ 2x + y - 4 = 4\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ \ }\left\{ \begin{matrix} x = 3 \\ y = 2 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x - y + 3 = - 4\ \ \\ 2x + y - 4 = - 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\left\{ \begin{matrix} x = - \frac{7}{3} \\ y = \frac{14}{3}\ \\ \end{matrix} \right.\ \]
\[Ответ:(6;8);(3;2).\]
