Решите уравнение: x^4+3x^3+4x^2+3x+1=0.

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

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

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

\[t^{2} + 2 + 3t = 0\]

\[t^{2} + 3t + 2 = 0\]

\[D = 3^{2} - 4 \cdot 1 \cdot 2 = 9 - 8 = 1\]

\[t_{1} = \frac{- 3 + \sqrt{1}}{2} = \frac{- 3 + 1}{2} = \frac{- 2}{2} =\]

\[= - 1\]

\[t_{2} = \frac{- 3 - \sqrt{1}}{2} = \frac{- 3 - 1}{2} = \frac{- 4}{2} =\]

\[= - 2\]

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

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

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

\[= - 3 < 0 \Longrightarrow нет\ решения.\]

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

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

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

\[x + 1 = 0\]

\[x = - 1\]

\[Ответ:\ - 1.\]