Решите уравнение: 2x^4-x^3-6x^2-x+2=0.

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

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

\[t = x + \frac{1}{x}\]

\[2t^{2} - 10 - t = 0\]

\[2t^{2} - t - 10 = 0\]

\[D = ( - 1)^{2} - 4 \cdot 2 \cdot ( - 10) =\]

\[= 1 + 80 = 81\]

\[t_{1} = \frac{1 + \sqrt{81}}{2 \cdot 2} = \frac{1 + 9}{4} = \frac{10}{4} = \frac{5}{2}\]

\[t_{2} = \frac{1 - \sqrt{81}}{2 \cdot 2} = \frac{1 - 9}{4} = \frac{- 8}{4} =\]

\[= - 2\]

\[1)\ x + \frac{1}{x} = \frac{5}{2}\ \ \ \ \ \ \ \ \ \ \ | \cdot 2x\]

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

\[D = ( - 5)^{2} - 4 \cdot 2 \cdot 2 =\]

\[= 25 - 16 = 9\]

\[x_{1} = \frac{5 + \sqrt{9}}{2 \cdot 2} = \frac{5 + 3}{4} = \frac{8}{4} = 2\]

\[x_{2} = \frac{5 - \sqrt{9}}{2 \cdot 2} = \frac{5 - 3}{4} = \frac{2}{4} = 0,5\]

\[2)\ x + \frac{1}{x} = - 2\ \ \ \ \ \ \ \ \ \ \ | \cdot x\]

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

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

\[x + 1 = 0\]

\[x = - 1\]

\[Ответ:\ 2;\ 0,5;\ - 1.\]