ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 943

\[1)\ \left\{ \begin{matrix} \frac{x^{4}}{y^{2}} + xy = 72 \\ \frac{y^{4}}{x^{2}} + xy = 9\ \ \\ \end{matrix}\text{\ \ }(:) \right.\ \]

\[\frac{x^{4} + xy^{3}}{y^{2}}\ :\frac{y^{4} + x^{3}y}{x^{2}} = \frac{72}{9}\]

\[\frac{x\left( x^{3} + y^{3} \right) \bullet x^{2}}{y\left( y^{3} + x^{3} \right) \bullet y^{2}} = 8\]

\[\left( \frac{x}{y} \right)^{3} = 8\]

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

\[x = 2y.\]

\[\frac{y^{4}}{x^{2}} + xy = 9\]

\[\frac{y^{4}}{(2y)^{2}} + y \bullet 2y = 9\]

\[\frac{y^{2}}{4} + 2y^{2} = 9\]

\[y^{2} + 8y^{2} = 36\]

\[9y^{2} = 36\]

\[y^{2} = 4\]

\[y = \pm 2;\]

\[x = 2 \bullet ( \pm 2) = \pm 4.\]

\[Ответ:\ \ ( - 4;\ - 2);\ (4;\ 2).\]

\[2)\ \left\{ \begin{matrix} \frac{x^{3}}{2y} + 3xy = 25 \\ \frac{y^{3}}{x} - 2xy = 16 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\left\{ \begin{matrix} \frac{x^{3}}{2y} = 25 - 3xy \\ \frac{y^{3}}{x} = 16 + 2xy \\ \end{matrix}\text{\ \ }( \bullet ) \right.\ \ \]

\[\frac{x^{3}}{2y} \bullet \frac{y^{3}}{x} = (25 - 3xy)(16 + 2xy)\]

\[\frac{\left( \text{xy} \right)^{2}}{2} = 400 + 50xy - 48xy - 6\left( \text{xy} \right)^{2}\]

\[6,5\left( \text{xy} \right)^{2} - 2xy - 400 = 0\]

\[13\left( \text{xy} \right)^{2} - 4\left( \text{xy} \right) - 800 = 0\]

\[D = 16 + 41\ 600 = 41\ 616\]

\[\left( \text{xy} \right)_{1} = \frac{4 - 204}{2 \bullet 13} = - \frac{100}{13};\]

\[\left( \text{xy} \right)_{2} = \frac{4 + 204}{2 \bullet 13} = 8;\]

\[xy = 8\ \ \ \]

\[y = \frac{8}{x}.\]

\[\frac{x^{3}}{2y} + 3xy = 25\]

\[\frac{x^{3}}{2}\ :\frac{8}{x} + 3x \bullet \frac{8}{x} = 25\]

\[\frac{x^{4}}{16} + 24 = 25\]

\[\frac{x^{4}}{16} = 1\]

\[x^{4} = 16\]

\[x = \pm 2;\]

\[y = \frac{8}{\pm 2} = \pm 4.\]

\[Ответ:\ \ ( - 2;\ - 4);\ (2;\ 4).\]