Решите уравнение: ||x^2+4x+1|-1|=2.

\[\left| \left| x^{2} + 4x + 1 \right| - 1 \right| = 2\]

\[\left| x^{2} + 4x + 1 \right| - 1 = 2\]

\[\left| x^{2} + 4x + 1 \right| = 3\]

\[x^{2} + 4x + 1 = 3\]

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

\[D = 4^{2} - 4 \cdot ( - 2) = 16 + 8 = 24\]

\[x_{1} = \frac{- 4 + \sqrt{24}}{2} = \frac{- 4 + 2\sqrt{6}}{2} =\]

\[= - 2 + \sqrt{6}\]

\[x_{2} = \frac{- 4 - \sqrt{24}}{2} = \frac{- 4 - 2\sqrt{6}}{2} =\]

\[= - 2 - \sqrt{6}\]

\[x^{2} + 4x + 1 = - 3\]

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

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

\[x + 2 = 0\]

\[x = - 2.\]

\[|x^{2} + 4x + 1| - 1 = - 2\]

\[\left| x^{2} + 4x + 1 \right| = - 1 \Longrightarrow\]

\[\Longrightarrow нет\ решения.\]

\[Ответ:\ - 2 + \sqrt{6}\ ;\ - 2 - \sqrt{6};\ - 2.\]