Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 366

\[\boxed{\mathbf{366}.}\]

\[1)\ \left\{ \begin{matrix} xy - \frac{x}{y} = 6\ \ \ \\ xy - \frac{y}{x} = \frac{15}{2} \\ \end{matrix} \right.\ ( - )\]

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

\[\frac{- x^{2} + y^{2}}{\text{xy}} = - \frac{3}{2}\ \ \ \ | \cdot ( - 2xy)\ \]

\[2x^{2} - 2y^{2} = 3xy\]

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

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

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

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

\[x = 2y:\]

\[2y^{2} = 8\]

\[y^{2} = 4\]

\[y = \pm 2.\]

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

\[y^{2} = - 11\]

\[нет\ корней.\]

\[y = 2;\ \ x = 2y = 4.\]

\[y = - 2;\ \ \ x = 2y = - 4.\]

\[Ответ:( - 4; - 2);\ \ (4;2).\]

\[2)\ \left\{ \begin{matrix} xy - 4 = \frac{7x}{y} \\ xy - \frac{3}{2} = \frac{y}{2x} \\ \end{matrix} \right.\ ( - )\]

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

\[\frac{14x^{2} - y^{2}}{2xy} = - \frac{5}{2}\ \ \ \ \ | \cdot 2xy\]

\[14x^{2} - y^{2} = - 5xy\]

\[14x^{2} + 5xy - y^{2} = 0\]

\[D = 25y^{2} + 56y^{2} = 81y^{2}\]

\[x_{1} = \frac{- 5y + 9y}{28} = \frac{y}{7};\ \ \]

\[\ x_{2} = \frac{- 5y - 9y}{28} = - \frac{y}{2}.\]

\[1)\ x = \frac{y}{7}:\]

\[\frac{y}{7} \cdot y = 7 \cdot \frac{y}{7} \cdot \frac{1}{y} + 4\]

\[\frac{y^{2}}{7} = 5\]

\[y^{2} = 35\]

\[y = \pm \sqrt{35}.\]

\[x = \frac{\sqrt{35}}{7};\ \ \ x = - \frac{\sqrt{35}}{7}.\]

\[2)\ x = - \frac{y}{2}\]

\[- \frac{y}{2} \cdot y = 7 \cdot \left( - \frac{y}{2} \right) \cdot \frac{1}{y} + 4\]

\[- \frac{y^{2}}{2} = 0,5\]

\[y^{2} = - 1\]

\[нет\ корней.\]

\[Ответ:\left( \frac{\sqrt{35}}{7};\sqrt{35} \right);\ \]

\[\ \left( - \frac{\sqrt{35}}{7}; - \sqrt{35} \right).\]