Решите уравнение: (x^2-2x)^2+12*(x-1)^2-1=0.

\[\left( x^{2} - 2x \right)^{2} + 12 \bullet (x - 1)^{2} - 1 =\]

\[= 0\]

\[(x - 1)^{2}\left( x^{2} - 2x - 1 + 12 \right) = 0\]

\[(x - 1)^{2}\left( x^{2} - 2x + 11 \right) = 0\]

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

\[x - 1 = 0\]

\[x = 1.\]

\[2)\ x² - 2x + 11 = 0\]

\[D = ( - 2)^{2} - 4 \cdot 1 \cdot 11 =\]

\[= 4 - 44 = - 40 < 0 \Longrightarrow\]

\[\Longrightarrow нет\ решения.\]

\[Ответ:x = 1.\]