Решите уравнение: |x^2-3x+1|=|x^2+5x-5|.

\[\left| x^{2} - 3x + 1 \right| = \left| x^{2} + 5x - 5 \right|\]

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

\[5x + 3x = 1 + 5\]

\[8x = 6\]

\[x = \frac{6}{8} = \frac{3}{4}.\]

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

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

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

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

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

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

\[x_{1} = \frac{- 1 + \sqrt{9}}{2} = \frac{- 1 + 3}{2} = \frac{2}{2} = 1\]

\[x_{2} = \frac{- 1 - \sqrt{9}}{2} = \frac{- 1 - 3}{2} = \frac{- 4}{2} =\]

\[= - 2\]

\[Ответ:\ \frac{3}{4};\ 1;\ \ - 2.\]