Решите уравнение (x^2+27x-57)^2=(x^2-3x+1)^2.

\[\left( x^{2} + 27x - 57 \right)^{2} = \left( x^{2} - 3x + 1 \right)^{2}\]

\[1)\ x^{2} + 27x - 57 = x^{2} - 3x + 1\]

\[27x + 3x = 1 + 57\]

\[30x - 58 = 0\]

\[2 \cdot (15x - 29) = 0\]

\[15x = 29\]

\[x = \frac{29}{15}.\]

\[2)\ x^{2} + 27x - 57 = - x^{2} + 3x - 1\]

\[2x^{2} + 24x - 56 = 0\]

\[2 \cdot \left( x^{2} + 15x - 28 \right) = 0\]

\[2 \cdot (x - 2)(x + 14) = 0\]

\[x = 2;\ \ \ x = - 14\]

\[Ответ:x = \frac{29}{15};\ \ x = 2;\ \ x = - 14.\]