Решите уравнение: (x^2+2x)^2+13*(x+1)^2-1=0.

\[\left( x^{2} + 2x \right)^{2} + 13 \bullet (x + 1)^{2} - 1 =\]

\[= 0\]

\[(x + 1)^{2}\left( x^{2} + 2x - 1 + 13 \right) = 0\]

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

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

\[1)\ x + 1 = 0\]

\[x = - 1.\]

\[2)\ x^{2} + 2x + 12 = 0\]

\[D = 2^{2} - 4 \cdot 1 \cdot 12 = 4 - 48 =\]

\[= - 44 < 0 \Longrightarrow нет\ решения.\]

\[Ответ:\ x = - 1.\]