ГДЗ по алгебре и начала математического анализа 10 класс Колягин Задание 323

\[\boxed{\mathbf{323}.}\]

\[1)\ 2x^{5} - x^{4} + 2x - 1 = 0\]

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

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

\[x^{4} + 1 = 0\ \ \ \ \ \ \ \ \ \ 2x - 1 = 0\]

\[нет\ корней\ \ \ \ \ \ \ \ x = 0,5.\]

\[Ответ:x = 0,5.\]

\[2)\ 4x^{5} - x^{3} - 4x^{2} + 1 = 0\]

\[x^{3}\left( 4x^{2} - 1 \right) - \left( 4x^{2} - 1 \right) = 0\]

\[\left( 4x^{2} - 1 \right)\left( x^{3} - 1 \right) = 0\]

\[4x^{2} - 1 = 0\ \ \ \ \ \ \ \ x^{3} - 1 = 0\]

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

\[x^{2} = \frac{1}{4}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x = 1\]

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

\[Ответ:x = \pm 0,5;\ \ x = 1.\]

\[3)\ 6x^{6} - x^{5} - x^{4} + 6x^{2} -\]

\[- x - 1 = 0\]

\[x^{4}\left( 6x^{2} - x - 1 \right) + 6x^{2} -\]

\[- x - 1 = 0\]

\[\left( 6x^{2} - x - 1 \right)\left( x^{4} + 1 \right) = 0\]

\[6x^{2} - x - 1 = 0\]

\[D = 1 + 24 = 25\]

\[x_{1} = \frac{1 + 5}{12} = \frac{1}{2};\ \ \ \]

\[x_{2} = \frac{1 - 5}{12} = - \frac{1}{3}.\]

\[x^{4} + 1 = 0\]

\[x^{4} = - 1\]

\[нет\ корней.\]

\[Ответ:\ \ x = 0,5;\ x = - \frac{1}{3}.\]

\[4)\ 4x^{6} + 4x^{5} - x^{4} - 5x^{3} -\]

\[- 4x^{2} + x + 1 = 0\]

\[4x^{5}(x + 1) - x^{3}(x + 1) -\]

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

\[(x + 1)\left( 4x^{5} - x^{3} - 4x^{2} + 1 \right) = 0\]

\[(x + 1)\left( x^{3}\left( 4x^{2} - 1 \right) - \left( 4x^{2} - 1 \right) \right) = 0\]

\[(x + 1)\left( 4x^{2} - 1 \right)\left( x^{3} - 1 \right) = 0\]

\[4x^{2} - 1 = 0\ \ \ \ \ \ x^{3} - 1 = 0\ \ \ \ \ \ \ \]

\[\ x + 1 = 0\]

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

\[\ \ x = - 1\]

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

\[Ответ:x = \pm 1;\ \ x = - 0,5.\]