Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 377

\[\boxed{\mathbf{377}.}\]

\[1)\ \left\{ \begin{matrix} \left( x^{2} + y^{2} \right)(x - y) = 13 \\ \text{xy}(x - y) = 6\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ (\ :)\]

\[\frac{x^{2} + y^{2}}{\text{xy}} = \frac{13}{6}\]

\[13xy = 6x^{2} + 6y^{2}\]

\[6x^{2} - 13xy + 6y^{2} = 0\]

\[D = 169y^{2} - 144y^{2} = 25y^{2}\]

\[x_{1} = \frac{13y - 5y}{12} = \frac{2y}{3};\ \ \ \]

\[x_{2} = \frac{13y + 5y}{12} = \frac{18y}{12} = \frac{3y}{2}.\]

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

\[\frac{2y^{2}}{3}\left( \frac{2y}{3} - y \right) = 6\]

\[- \frac{2y^{3}}{9} = 6\]

\[- 2y^{3} = 54\]

\[y^{3} = - 27\]

\[y = - 3.\]

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

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

\[\frac{3y^{2}}{2} \cdot \left( \frac{3y}{2} - y \right) = 6\]

\[\frac{3y^{3}}{4} = 6\]

\[3y^{3} = 24\]

\[y^{3} = 8\]

\[y = 2.\]

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

\[Ответ:( - 2;\ - 3);(3;2).\]

\[2)\ \left\{ \begin{matrix} 4 \cdot (x + y)\left( x^{2} - xy + y^{2} \right) = 9x^{2}y^{2} \\ 4 \cdot \left( x^{2} + y^{2} \right) = 9x^{2}y^{2} - 8xy\ \ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Пусть\ x + y = a;\ \ xy = b;\]

\[(x + y) = a^{2};\]

\[x^{2} + 2xy + y^{2} = a^{2};\]

\[x^{2} + y^{2} = a^{2} - 2b.\]

\[\left\{ \begin{matrix} 4a\left( a^{2} - 3b \right) = 9b^{2}\text{\ \ \ \ \ \ \ \ \ \ \ } \\ 4 \cdot \left( a^{2} - 2b \right) = 9b^{2} - 8b \\ \end{matrix} \right.\ \text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} 4a^{3} - 12ab = 9b^{2}\text{\ \ \ \ \ } \\ 4a^{2} - 8b = 9b^{2} - 8b \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 9b^{2} = 4a^{3} - 12ab \\ 9b^{2} = 4a^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} 4a^{3} - 12ab = 9b^{2} \\ b^{2} = \frac{4}{9}a^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[1)\ \left\{ \begin{matrix} b = \frac{2}{3}\text{a\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4a^{3} - 12a \cdot \frac{2}{3}a = 9 \cdot \frac{4}{9}a^{2} \\ \end{matrix} \right.\ \]

\[4a^{3} - 8a^{2} - 4a^{2} = 0\]

\[4a^{3} - 12a^{2} = 0\]

\[4a^{2}(a - 3) = 0\]

\[a = 0;\ \ a = 3\]

\[b = 0;\ \ \ b = 2.\]

\[2)\ \left\{ \begin{matrix} b = - \frac{2}{3}\text{a\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 4a^{3} - 12a \cdot \left( - \frac{2}{3} \right)a = 9 \cdot \frac{4}{9}a^{2} \\ \end{matrix} \right.\ \]

\[4a^{3} + 8a^{2} - 4a^{2} = 0\]

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

\[4a^{2}(a + 1) = 0\]

\[a = 0;\ \ \ \ a = - 1\]

\[b = 0;\ \ \ b = \frac{2}{3}.\]

\[Обратная\ замена:\]

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

\[\text{\ \ \ }\left\{ \begin{matrix} x = 0 \\ y = 0 \\ \end{matrix} \right.\ ;\]

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

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

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

\[y_{1} + y_{2} = 3;\ \ \ y_{1} \cdot y_{2} = 2\]

\[y_{1} = 2;\ \ \ \ \ \ \ \ \ \ \ \ \ y_{2} = 1\]

\[x_{1} = 3 - 2 = 1;\ \ \ \]

\[x_{2} = 3 - 1 = 2.\]

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

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

\[- x^{2} - x - \frac{2}{3} = 0\ \ \ \ | \cdot ( - 3)\]

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

\[D = 9 - 24 = - 15 < 0\]

\[нет\ корней.\]

\[Ответ:(0;0);(1;2);(2;1).\]