Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 368

\[\boxed{\mathbf{368}.}\]

\[1)\ \left\{ \begin{matrix} (x - y)\left( x^{2} + y^{2} \right) = 65 \\ (x + y)\left( x^{2} - y^{2} \right) = 5\ \ \ \\ \end{matrix} \right.\ \ (\ :)\]

\[\frac{(x - y)\left( x^{2} + y^{2} \right)}{(x + y)\left( x^{2} - y^{2} \right)} = 13\]

\[\frac{x^{2} + y^{2}}{(x + y)^{2}} = 13\]

\[x^{2} + y^{2} = 13 \cdot (x + y)^{2}\]

\[x^{2} + y^{2} = 13x^{2} + 26xy + 13y^{2}\]

\[12x^{2} + 26xy + 12y^{2} = 0\]

\[D_{1} = 169y^{2} - 144y^{2} = 25y^{2}\]

\[x_{1} = \frac{- 13y + 5y}{12} = - \frac{8y}{12} = - \frac{2y}{3};\]

\[x_{2} = \frac{- 13y - 5y}{12} = - \frac{18y}{12} =\]

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

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

\[\left( - \frac{2y}{3} + y \right)\left( \frac{4y^{2}}{9} - y^{2} \right) = 5\]

\[\frac{y}{3} \cdot \left( - \frac{5y^{2}}{9} \right) = 5\ \ \ \ \ \ \ | \cdot \frac{27}{5}\]

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

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

\[y = - 3.\]

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

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

\[\left( - \frac{3y}{2} + y \right)\left( \frac{9}{4}y^{2} - y^{2} \right) = 5\]

\[- \frac{y}{2} \cdot \frac{5y^{2}}{4} = 5\ \ \ \ | \cdot \left( - \frac{8}{5} \right)\]

\[y^{3} = - 8\]

\[y = - 2.\]

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

\[Ответ:(2;\ - 3);(3; - 2).\]

\[2)\ \left\{ \begin{matrix} x^{3} + 4y = y^{3} + 16x \\ 1 + y^{2} = 5 \cdot \left( 1 + x^{2} \right) \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y^{3} - 4y = x^{3} - 16x \\ y^{2} - 4 = 5x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} y\left( y^{2} - 4 \right) = x\left( x^{2} - 16 \right) \\ \left( y^{2} - 4 \right) = 5x^{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \ (\ :)\text{\ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = \frac{x^{2} - 16}{5x}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ 1 + \left( \frac{x^{2} - 16}{5x} \right)^{2} = 5 + 5x^{2} \\ \end{matrix} \right.\ \]

\[\frac{\left( x^{2} - 16 \right)^{2}}{25x^{2}} = 5x^{2} + 4\]

\[x^{4} - 32x^{2} + 256 =\]

\[= 125x^{4} + 100x^{2}\]

\[124x^{4} + 132x^{2} - 256 = 0\ \ \ \ |\ :4\]

\[31x^{4} + 33x^{2} - 64 = 0\]

\[{x^{2} = t \geq 0: }{31t^{2} + 33t - 64 = 0}\]

\[D = 1089 + 7936 = 9025\]

\[t_{1} = \frac{- 33 + 95}{62} = 1;\ \ \]

\[\ t_{2} = \frac{- 33 - 95}{62} =\]

\[= - \frac{128}{62}\ (не\ подходит).\]

\[x^{2} = 1\]

\[x = \pm 1.\]

\[x = 1:\]

\[y = \frac{1 - 16}{5} = - 3.\]

\[x = - 1:\]

\[y = \frac{1 - 16}{- 5} = 3.\]

\[Ответ:(1;\ - 3);( - 1;3).\]