ГДЗ по алгебре и начала математического анализа 11 класс Колягин Задание 944

\[1)\ \left\{ \begin{matrix} \frac{x}{y} + x^{4}y = \frac{1}{xy^{2}} + x^{2} \\ \frac{1}{x} + x^{2}y^{2} + 4y^{2} = 0 \\ \end{matrix} \right.\ \]

\[\frac{1}{x} + x^{2}y^{2} + 4y^{2} = 0\]

\[\frac{1}{x}\left( 1 + x^{3}y^{2} \right) = - 4y^{2}\]

\[1 + x^{3}y^{2} = - 4xy^{2}.\]

\[\frac{x}{y} + x^{4}y = \frac{1}{xy^{2}} + x^{2}\]

\[\frac{x}{y}\left( 1 + x^{3}y^{2} \right) = \frac{1}{xy^{2}} + x^{2}\]

\[\frac{x}{y} \bullet \left( - 4xy^{2} \right) = \frac{1}{xy^{2}} + x^{2}\]

\[- 4x^{2}y = \frac{1}{xy^{2}} + x^{2}\]

\[- 4x^{3}y^{3} = 1 + x^{3}y^{2}\]

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

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

\[y = \frac{1}{x^{2}}.\]

\[\frac{1}{x} + x^{2}y^{2} + 4y^{2} = 0\]

\[\frac{1}{x} + x^{2} \bullet \frac{1}{x^{4}} + 4 \bullet \frac{1}{x^{4}} = 0\]

\[\frac{1}{x} + \frac{1}{x^{2}} + \frac{4}{x^{4}} = 0\]

\[x^{3} + x^{2} + 4 = 0\]

\[(x + 2)\left( x^{2} - x + 2 \right) = 0\]

\[x + 2 = 0\]

\[x = - 2;\]

\[y = \frac{1}{( - 2)^{2}} = \frac{1}{4}.\]

\[Ответ:\ \ \left( - 2;\ \frac{1}{4} \right).\]

\[2)\ \left\{ \begin{matrix} x + \frac{1}{x^{3}y^{3}} = x^{3}y + \frac{1}{xy^{2}} \\ \frac{1}{x} + x^{3}y^{3} + 10y^{2} = 0\ \ \\ \end{matrix} \right.\ \]

\[x + \frac{1}{x^{3}y^{3}} = x^{3}y + \frac{1}{xy^{2}}\]

\[\frac{x^{4}y^{3} + 1 - x^{6}y^{4} - x^{2}y}{x^{3}y^{3}} = 0\]

\[\frac{x^{4}y^{3}\left( 1 - x^{2}y \right) + \left( 1 - x^{2}y \right)}{x^{3}y^{3}} = 0\]

\[\frac{\left( x^{4}y^{3} + 1 \right)\left( 1 - x^{2}y \right)}{x^{3}y^{3}} = 0\]

\[y_{1} = - \frac{1}{\sqrt[3]{x^{4}}};\ \ \ y_{2} = \frac{1}{x^{2}}.\]

\[1)\ \frac{1}{x} + x^{3} \bullet \left( - \frac{1}{x^{4}} \right) + 10 \bullet \frac{1}{\sqrt[3]{x^{8}}} = 0\]

\[\frac{1}{x} - \frac{1}{x} + \frac{10}{\sqrt[3]{x^{8}}} = 0\]

\[\frac{10}{\sqrt[3]{x^{8}}} = 0\]

\[x \in \varnothing.\]

\[2)\ \frac{1}{x} + x^{3} \bullet \frac{1}{x^{6}} + 10 \bullet \frac{1}{x^{4}} = 0\]

\[\frac{1}{x} + \frac{1}{x^{3}} + \frac{10}{x^{4}} = 0\]

\[x^{3} + x + 10 = 0\]

\[(x + 2)\left( x^{2} - 2x + 5 \right) = 0\]

\[x + 2 = 0\]

\[x = - 2;\]

\[y = \frac{1}{( - 2)^{2}} = \frac{1}{4}.\]

\[Ответ:\ \ \left( - 2;\ \frac{1}{4} \right).\]