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

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

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

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

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

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

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

\[= 1\]

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

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

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

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

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

\[x - 1 = 0\]

\[x = 1.\]

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

\[x² - x + 1 = 0\]

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

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

\[Ответ:1.\]