Решите уравнение: (x^2-x+1)^2-2x^2+2x=5.

Ответ:

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

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

\[x^{3}(x - 2) + (x - 2)(x + 2) = 0\]

\[(x - 2)\left( x^{3} + x + 2 \right) = 0\]

\[1)\ x - 2 = 0\]

\[x = 2.\]

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

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

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

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

\[x + 1 = 0\]

\[x = - 1.\]

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

\[D = 1 - 8 = - 7 < 0\]

\[нет\ корней.\]

\[Ответ:x = 2;x = - 1.\ \]