ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 620

\[\boxed{\mathbf{620}.}\]

\[1)\ \left\{ \begin{matrix} \sqrt[3]{\frac{y}{x}} - 2\sqrt[3]{\frac{x}{y}} = 1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \sqrt{x + y} + \sqrt{x - y + 11} = 5 \\ \end{matrix} \right.\ \]

\[заменим\ t = \sqrt[3]{\frac{y}{x}}:\]

\[t - 2 \cdot \frac{1}{t} = 1\]

\[t^{2} - t - 2 = 0\]

\[t_{1} + t_{2} = 1;\ \ \ t_{1} \cdot t_{2} = - 2\]

\[t_{1} = 2;\ \ \ t_{2} = - 1.\]

\[1)\ \ \sqrt[3]{\frac{y}{x}} = 2\]

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

\[y = 8x.\]

\[\sqrt{x + y} + \sqrt{x - y + 11} = 5\]

\[\sqrt{x + 8x} + \sqrt{x - 8x + 11} = 5\]

\[\left( \sqrt{9x} + \sqrt{11 - 7x} \right)^{2} = 25\]

\[9x + 2\sqrt{9x(11 - 7x)} +\]

\[+ 11 - 7x = 25\]

\[2\sqrt{99x - 63x^{2}} = 14 - 2x\ \ \ |\ :2\]

\[\sqrt{99x - 63x^{2}} = 7 - x\]

\[99x - 63x^{2} = (7 - x)^{2}\]

\[99x - 63x^{2} = 49 - 14x + x^{2}\]

\[64x^{2} - 113x + 49 = 0\]

\[D = 12\ 769 - 12\ 544 = 225\]

\[x_{1} = \frac{113 + 15}{128} = 1;\ \ \ \]

\[x_{2} = \frac{113 - 15}{128} = \frac{98}{128} = \frac{49}{64}.\]

\[y_{1} = 8 \cdot 1 = 8;\ \ \]

\[y_{2} = 8 \cdot \frac{49}{64} = \frac{49}{8} = 6\frac{1}{8}.\]

\[2)\ \sqrt[3]{\frac{y}{x}} = - 1\]

\[\frac{y}{x} = - 1\]

\[y = - x.\]

\[\sqrt{x + y} + \sqrt{x - y + 11} = 5\]

\[\sqrt{x - x} + \sqrt{x + x + 11} = 5\]

\[\left( \sqrt{2x + 11} \right)^{2} = 5^{2}\]

\[2x + 11 = 25\]

\[2x = 14\]

\[x = 7.\]

\[y = - x = - 7.\]

\[Ответ:(1;8);\left( \frac{49}{64};6\frac{1}{8} \right);(7; - 7).\]

\[2)\ \left\{ \begin{matrix} \sqrt{\frac{x}{y}} - \sqrt{\frac{y}{x}} = \frac{3}{2}\text{\ \ \ } \\ x + y + xy = 9 \\ \end{matrix} \right.\ \]

\[Заменим\ t = \sqrt{\frac{x}{y}}:\]

\[t - \frac{1}{t} = \frac{3}{2}\ \ \ | \cdot 2t\]

\[2t^{2} - 3t - 2 = 0\]

\[D = 9 + 16 = 25\]

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

\[\ t_{2} = \frac{3 - 5}{4} = - \frac{1}{2}.\]

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

\[\frac{x}{y} = 4\]

\[x = 4y.\]

\[x + y + xy = 9\]

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

\[4y^{2} + 5y - 9 = 0\]

\[D = 25 + 144 = 169\]

\[y_{1} = \frac{- 5 + 13}{8} = 1;\ \]

\[\ y_{2} = \frac{- 5 - 13}{8} = - \frac{9}{4} = - 2,25.\]

\[x_{1} = 4y = 4 \cdot 1 = 4;\ \]

\[\ x_{2} = 4 \cdot \left( - \frac{9}{4} \right) = - 9.\]

\[2)\ \sqrt{\frac{x}{y}} = - \frac{1}{2}\]

\[\frac{x}{y} = \frac{1}{4}\]

\[y = 4x.\]

\[x + y + xy = 9\]

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

\[4x^{2} + 5x - 9 = 0\]

\[D = 25 + 144 = 169\]

\[x_{1} = \frac{- 5 - 13}{8} = - \frac{9}{4};\ \ \]

\[x_{2} = \frac{- 5 + 13}{8} = 1.\]

\[y_{1} = 4;\ \ y_{2} = - 9.\]

\[Корни\ не\ подходят\ \]

\[по\ проверке.\]

\[Ответ:(4;1);( - 9; - 2,25).\]