Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 342

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\[x^{3} + 3\sqrt{2}x^{2} + 5x + 12 = 0\]

\[Пусть\ x_{1}x_{2}x_{3} - корни;\]

\[\text{\ \ }x_{1} \cdot x_{2} = 1.\]

\[\left\{ \begin{matrix} x_{1} + x_{2} + x_{3} = - 3\sqrt{2}\text{\ \ \ \ } \\ x_{1}x_{2} + x_{2}x_{3} + x_{1}x_{3} = 5 \\ x_{1}x_{2}x_{3} = - \sqrt{2}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ } \\ \end{matrix} \right.\ \]

\[Так\ как\ x_{1} \cdot x_{2} = 1:\]

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

\[\left\{ \begin{matrix} 1 - \sqrt{2}x_{2} - \sqrt{2}x_{1} = 5 \\ x_{1} + x_{2} - \sqrt{2} = - 3\sqrt{2} \\ \end{matrix} \right.\ \]

\[\left\{ \begin{matrix} x_{1} + x_{2} = - 2\sqrt{2} \\ x_{1} \cdot x_{2} = 1\ \ \ \ \ \ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[x_{1} = - 2\sqrt{2} - x_{2}\]

\[\left( - 2\sqrt{2} - x_{2} \right)x_{2} = 1\]

\[x_{2}^{2} + 2\sqrt{2x_{2}} + 1 = 0\]

\[D = 8 - 4 = 4\]

\[x_{1} = \frac{- 2\sqrt{2} + 2}{2} = - \sqrt{2} + 1;\ \]

\[\ x_{2} = \frac{- 2\sqrt{2} - 2}{2} = - \sqrt{2} - 1.\]

\[Ответ:x = - 1 - \sqrt{2};\ - \sqrt{2};\]

\[\ - \sqrt{2} + 1.\]