ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 319

\[\boxed{\mathbf{319}.}\]

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

\[= 2x\left( x^{3} + 3 \right) - 5\]

\[2x^{4} + 2x + x^{3} + 1 + x^{2} =\]

\[= 2x^{4} + 6x - 5\]

\[x^{3} + x^{2} - 4x + 6 = 0\]

\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ \pm 6.\]

\[P(x) =\]

\[= (x + 3)\left( x^{2} - 2x + 2 \right) = 0\]

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

\[D_{1} = 1 - 2 = - 1 < 0\]

\[нет\ корней.\]

\[Ответ:x = - 3.\]

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

\[= (x + 1)^{2} + 16x^{2} - 6\]

\[4x^{4} - 4x^{2} + 1 + 4x^{3} - 4x^{2} +\]

\[+ x = x^{2} + 2x + 1 + 16x^{2} - 6\]

\[4x^{4} + 4x^{3} - 25x^{2} - x + 6 = 0\]

\[Делители:\ \pm 1;\ \pm 2;\ \pm 3;\ \pm 6.\]

\[P(x) = (x - 2)(x + 3)\left( 4x^{2} - 1 \right) =\]

\[= (x - 2)(x + 3)(2x - 1)(2x + 1) = 0\]

\[Ответ:\ x = - 3;\ \ x = 2;\ \ \]

\[x = \pm 0,5.\]