ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 362

\[\boxed{\mathbf{362}.}\]

\[1)\ \left\{ \begin{matrix} x - y = 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{4} - 3yx^{2} - 4y^{2} = 0 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[\ \left\{ \begin{matrix} y = x - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ x^{4} - 3x^{2}(x - 2) - 4 \cdot (x - 2)^{2} = 0 \\ \end{matrix} \right.\ \]

\[x^{4} - 3x^{3} + 6x^{2} - 4x^{2} +\]

\[+ 16x - 16 = 0\]

\[x^{4} - 3x^{3} + 2x^{2} + 16x - 16 = 0\]

\[\left( x^{2} + x - 2 \right)\left( x^{2} - 4x + 8 \right) = 0\]

\[(x - 1)(x + 2)\left( x^{2} - 4x + 8 \right) = 0\]

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

\[D_{1} = 4 - 8 = - 4 < 0\]

\[нет\ корней.\]

\[x = 1:\]

\[y = 1 - 2 = - 1;\]

\[x = - 2:\]

\[y = - 2 - 2 = - 4.\]

\[Ответ:(1;\ - 1);( - 2;\ - 4).\]

\[2)\ \left\{ \begin{matrix} x + y = - 2\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{4} - 5x^{2}y + 3y^{2} = 0 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} y = 2 - x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \\ 2x^{4} - 5x^{2}(2 - x) + 3 \cdot (2 - x)^{2} = 0 \\ \end{matrix} \right.\ \]

\[2x^{4} - 10x^{2} + 5x^{3} + 12 -\]

\[- 12x + 3x^{2} = 0\]

\[2x^{4} + 5x^{3} - 7x^{2} -\]

\[- 12x + 12 = 0\]

\[(x - 1)(x + 2)\left( 2x^{2} + 3x - 6 \right) =\]

\[= 0\]

\[2x^{2} + 3x - 6 = 0\]

\[D = 9 + 48 = 57\]

\[x = \frac{- 3 \pm \sqrt{57}}{4}.\]

\[x = 1:\]

\[y = 2 - 1 = 1.\]

\[x = - 2:\]

\[y = 2 + 2 = 4.\]

\[x = \frac{- 3 + \sqrt{57}}{4}:\]

\[y = 2 - \frac{- 3 + \sqrt{57}}{4} = \frac{11 - \sqrt{57}}{4}.\]

\[x = \frac{- 3 - \sqrt{57}}{4}:\]

\[y = 2 + \frac{3 + \sqrt{57}}{4} = \frac{11 + \sqrt{57}}{4}.\]

\[Ответ:(1;1);( - 2;4);\]

\[\left( \frac{- 3 + \sqrt{57}}{4};\ \frac{11 - \sqrt{57}}{4} \right);\]

\[\left( \frac{- 3 - \sqrt{57}}{4};\frac{11 + \sqrt{57}}{4} \right).\]