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МатематикаРешебник по алгебре и начала математического анализа 10 класс Колягин Задание 369
Задание 369
\[\boxed{\mathbf{369}.}\]
\[1)\ \left\{ \begin{matrix} x + 2y = 3y^{2} \\ 2x + y = 3x^{2} \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]
\[\left\{ \begin{matrix} x = 3y^{2} - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 9y^{4} - 12y^{3} + 2y^{2} + y = 0 \\ \end{matrix} \right.\ \]
\[\left\{ \begin{matrix} x = 3y^{2} - 2y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y(y - 1)\left( 9y^{2} - 3y - 1 \right) = 0 \\ \end{matrix} \right.\ \]
\[y = 0:\]
\[x = 0.\]
\[y = 1:\]
\[x = 3 \cdot 1 - 2 = 1.\]
\[9y^{2} - 3y - 1 = 0\]
\[D = 9 + 36 = 45\]
\[y_{1,2} = \frac{3 \pm 3\sqrt{5}}{18} = \frac{1 \pm \sqrt{5}}{6}.\]
\[y = \frac{1 + \sqrt{5}}{6}:\]
\[x = 3 \cdot \left( \frac{1 + \sqrt{5}}{6} \right)^{2} -\]
\[- 2 \cdot \frac{1 + \sqrt{5}}{6} = \frac{1 - \sqrt{5}}{6}.\]
\[y = \frac{1 - \sqrt{5}}{6}:\]
\[x = 3 \cdot \left( \frac{1 - \sqrt{5}}{6} \right)^{2} -\]
\[- 2 \cdot \frac{1 - \sqrt{5}}{6} = \frac{1 + \sqrt{5}}{6}.\]
\[Ответ:(0;0);(1;1);\]
\[\left( \frac{1 + \sqrt{5}}{6};\frac{1 - \sqrt{5}}{6} \right);\]
\[\left( \frac{1 - \sqrt{5}}{6};\ \frac{1 + \sqrt{5}}{6} \right).\]
\[2)\ \left\{ \begin{matrix} x^{2} + x + xy = 8 \\ y^{2} + y + xy = 4 \\ \end{matrix} \right.\ ( + )\]
\[x^{2} + y^{2} + x + y + 2xy = 12\]
\[x^{2} + 2xy + y^{2} + x + y - 12 = 0\]
\[(x + y)^{2} + (x + y) - 12 = 0\]
\[x + y = t:\]
\[t^{2} + t - 12 = 0\]
\[t_{1} + t_{2} = - 1;\ \ t_{1} \cdot t_{2} = - 12\]
\[t_{1} = - 4;\ \ \ \ \ t_{2} = 3\]
\[1)\ \left\{ \begin{matrix} x + y = - 4\ \ \ \ \ \ \ \ \ \\ y^{2} + y + xy = 4 \\ \end{matrix} \right.\ \text{\ \ \ }\]
\[\ \left\{ \begin{matrix} x = - 4 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + y + y( - 4 - y) = 4 \\ \end{matrix} \right.\ \]
\[y^{2} + y - 4y - y^{2} = 4\]
\[- 3y = 4\]
\[y = - \frac{4}{3};\]
\[x = - 4 + \frac{4}{3} = - \frac{8}{3}.\]
\[2)\ \left\{ \begin{matrix} x + y = 3\ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + y + xy = 4 \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]
\[\ \left\{ \begin{matrix} x = 3 - y\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ y^{2} + y + y(3 - y) = 4 \\ \end{matrix} \right.\ \]
\[y^{2} + y + 3y - y^{2} = 4\]
\[4y = 4\]
\[y = 1.\]
\[x = 3 - 1 = 2.\]
\[Ответ:\left( - \frac{8}{3};\ - \frac{4}{3} \right);\ \ (2;1).\]
