Решите уравнение: (x^2-2x)^2+12*(x^2-2x)+11=0.

\[t = x^{2} - 2x\]

\[t^{2} + 12t + 11 = 0\]

\[D = 12^{2} - 4 \cdot 1 \cdot 11 =\]

\[= 144 - 44 = 100\]

\[t_{1} = \frac{- 12 + \sqrt{100}}{2} = \frac{- 12 + 10}{2} =\]

\[= \frac{- 2}{2} = - 1\]

\[t_{2} = \frac{- 12 - \sqrt{100}}{2} = \frac{- 12 - 10}{2} =\]

\[= \frac{- 22}{2} = - 11\]

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

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

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

\[x - 1 = 0\]

\[x = 1.\]

\[2)\ x^{2} - 2x = - 11\]

\[x^{2} - 2x + 11 = 0\]

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

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

\[\Longrightarrow нет\ решения.\]

\[Ответ:x = 1.\]