Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 400

\[\boxed{\mathbf{400}.}\]

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

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

\[2 \cdot \left( x^{2} + y^{2} \right) = 5xy\]

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

\[D = 25y^{2} - 16y^{2} = 9y^{2}\]

\[x_{1} = \frac{5y + 3y}{4} = 2y;\ \ \]

\[\ x_{2} = \frac{5y - 3y}{4} = \frac{y}{2}.\]

\[1)\ x = 2y:\]

\[4y^{2} + 2y^{2} = 6\]

\[6y^{2} = 6\]

\[y^{2} = 1\]

\[y = \pm 1.\]

\[x = \pm 2.\]

\[2)\ x = \frac{y}{2};\ \ \ y = 2x:\]

\[x^{2} + 8x^{2} = 6\]

\[9x^{2} = 6\]

\[x^{2} = \frac{6}{9}\]

\[x = \pm \frac{\sqrt{6}}{3}.\]

\[y = \pm \frac{2\sqrt{6}}{3}.\]

\[Ответ:( - 2; - 1);(2;1);\]

\[\left( \frac{\sqrt{6}}{3};\frac{2\sqrt{6}}{3} \right);\ \ \left( - \frac{\sqrt{6}}{3}; - \frac{2\sqrt{6}}{3} \right).\]

\[2)\ \left\{ \begin{matrix} \frac{1}{x} - \frac{1}{y} = \frac{1}{6}\text{\ \ \ \ \ \ \ \ \ } \\ xy^{2} - x^{2}y = 6 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\frac{1}{x} = \frac{1}{6} + \frac{1}{y}\]

\[\frac{1}{x} = \frac{y + 6}{6y}\]

\[x = \frac{6y}{y + 6}\]

\[\left\{ \begin{matrix} x = \frac{6y}{y + 6}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \frac{6y^{3}}{y + 6} - \frac{36y^{3}}{(y + 6)^{2}} = 6 \\ \end{matrix} \right.\ \]

\[\frac{6y^{3}}{y + 6} - \frac{36y^{3}}{(y + 6)^{2}} =\]

\[= 6\ \ \ \ \ | \cdot \left( \frac{(y + 6)^{2}}{6} \right)\]

\[y^{3}(y + 6) - 6y^{3} = (y + 6)^{2}\]

\[y^{4} + 6y^{3} - 6y^{3} =\]

\[= y^{2} + 12y + 36\]

\[y_{4} - y^{2} - 12y - 36 = 0\]

\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ldots; \pm 36.\]

\[P(y) = (y - 3)(y + 2)\left( y^{2} + y + 6 \right) = 0.\]

\[y^{2} + y + 6 = 0\]

\[D = 1 - 24 = - 23 < 0\]

\[нет\ корней.\]

\[y = 3:\]

\[x = \frac{6 \cdot 3}{3 + 6} = 2.\]

\[y = - 2:\]

\[x = \frac{- 6 \cdot 2}{- 2 + 6} = - \frac{12}{4} = - 3.\]

\[Ответ:( - 3; - 2);(2;3).\]

\[3)\ \left\{ \begin{matrix} x^{2} - xy + y^{2} = 21 \\ y^{2} - 2xy + 15 = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

\[\ \left\{ \begin{matrix} x^{2} - xy + y^{2} = 21 \\ x = \frac{y^{2} + 15}{2y}\text{\ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[x = \frac{y^{2} + 15}{2y}:\]

\[\frac{\left( y^{2} + 15 \right)^{2}}{4y^{2}} - \frac{y\left( y^{2} + 15 \right)}{2y} +\]

\[+ y^{2} = 21\ \ \ | \cdot 4y^{2}\]

\[y^{4} + 30y^{2} + 225 - 2y^{4} -\]

\[- 30y^{2} + 4y^{4} - 84y^{2} = 0\]

\[3y^{4} - 84y^{2} + 225 = 0\ \ \ \ \ |\ :3\]

\[y^{4} - 28y^{2} + 75 = 0\]

\[t = y^{2} \geq 0:\]

\[t^{2} - 28t + 75 = 0\]

\[D_{1} = 196 - 75 = 121\]

\[t_{1} = 14 + 11 = 25;\ \ \ \]

\[t_{2} = 14 - 11 = 3.\]

\[1)\ y^{2} = 25\]

\[y = \pm 5.\]

\[y = 5:\]

\[x = 4.\]

\[y = - 5:\]

\[x = - 4.\]

\[2)\ y^{2} = 3\]

\[y = \pm \sqrt{3}\]

\[y = \sqrt{3}:\]

\[x = 3\sqrt{3}.\]

\[y = - \sqrt{3}:\]

\[x = - 3\sqrt{3}.\]

\[Ответ:(4;5);( - 4; - 5);\left( 3;3\sqrt{3} \right);\]

\[\left( - 3; - 3\sqrt{3} \right).\]

\[4)\ \left\{ \begin{matrix} x^{2} - 3xy + 2y^{2} = 3\ \ \ | \cdot 2 \\ 2x^{2} - 2xy - y^{2} = - 6\ \ \ \ \ \ \ \\ \end{matrix} \right.\ \text{\ \ \ \ }\]

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

\[4x^{2} - 8xy + 3y^{2} = 0\]

\[D = 64y^{2} - 48y^{2} = 16y^{2}\]

\[x_{1} = \frac{8y - 4y}{8} = \frac{y}{2};\ \ \ \]

\[x_{2} = \frac{8y + 4y}{8} = \frac{3y}{2}\]

\[1)\ x = \frac{y}{2} \rightarrow y = 2x:\]

\[x^{2} - 3x \cdot 2x + 2 \cdot (2x)^{2} = 3\]

\[x^{2} - 6x^{2} + 8x^{2} = 3\]

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

\[x^{2} = 1\]

\[x = \pm 1.\]

\[x = 1 \rightarrow y = 2;\]

\[x = - 1 \rightarrow y = - 2.\]

\[2)\ x = \frac{3y}{2}:\]

\[(1,5y)^{2} - 3y(1,5y) + 2y^{2} = 3\]

\[2,25y^{2} - 4,5y^{2} + 2y^{2} = 3\]

\[- 0,25y^{2} = 3\]

\[нет\ корней.\]

\[Ответ:(1;2);( - 1; - 2).\]