Решите уравнение: |x^2-2x-1|=2.

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

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

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

\[D = ( - 2)^{2} - 4 \cdot ( - 3) =\]

\[= 4 + 12 = 16\]

\[x_{1} = \frac{2 + \sqrt{16}}{2} = \frac{2 + 4}{2} = \frac{6}{2} = 3\]

\[x_{2} = \frac{2 - \sqrt{16}}{2} = \frac{2 - 4}{2} = \frac{- 2}{2} =\]

\[= - 1\]

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

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

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

\[x - 1 = 0\]

\[x = 1.\]

\[Ответ:3;1;1.\]