ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 385

\[\boxed{\mathbf{385}.}\]

\[P(x) = x^{3} + ax^{2} + bx - 2 = 0;\ \ \]

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

\[P\left( 1 + \sqrt{3} \right) = \left( 1 + \sqrt{3} \right)^{3} +\]

\[+ a\left( 1 + \sqrt{3} \right)^{2} +\]

\[+ b\left( 1 + \sqrt{3} \right) - 2 = 0\]

\[1 + 3\sqrt{3} + 9 + 3\sqrt{3} + a +\]

\[+ 2a\sqrt{3} + 3a + b + b\sqrt{3} - 2 = 0\]

\[\left( 4 + 2\sqrt{3} \right)a + \left( 1 + \sqrt{3} \right)b +\]

\[+ 6\sqrt{3} + 8 = 0.\]

\[P\left( 1 - \sqrt{3} \right) = \left( 1 - \sqrt{3} \right)^{3} +\]

\[+ a\left( 1 - \sqrt{3} \right)^{2} + b\left( 1 - \sqrt{3} \right) -\]

\[- 2 = 0\]

\[1 - 3\sqrt{3} + 9 - 3\sqrt{3} + a -\]

\[- 2a\sqrt{3} + 3a +\]

\[+ \left( 1 - \sqrt{3} \right)b - 2 = 0\]

\[\left( 4 - 2\sqrt{3} \right)a + \left( 1 - \sqrt{3} \right)b -\]

\[- 6\sqrt{3} + 8 = 0.\]

\[8a + 2b + 16 = 0\]

\[b = - 8 - 4a:\]

\[\left( 4 + 2\sqrt{3} \right)a +\]

\[+ \left( 1 + \sqrt{3} \right)( - 8 - 4a) +\]

\[+ 6\sqrt{3} + 8 = 0\]

\[4a + 2\sqrt{3}a - 8 - 8\sqrt{3} - 4a -\]

\[- 4a\sqrt{3} + 6\sqrt{3} + 8 = 0\]

\[- 2\sqrt{3}a - 2\sqrt{3} = 0\]

\[a = - 1.\]

\[b = - 8 - 4 \cdot ( - 1) = - 4.\]

\[Получили:\]

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

\[\left( x - 1 - \sqrt{3} \right)\left( x - 1 + \sqrt{3} \right) =\]

\[= x^{2} - 2x - 2.\]

\[Ответ:x = - 1;\ \ a = - 1;\]

\[b = - 4.\]