Решите уравнение методом замены переменной: (x^2+3x+1)(x^2+3x+3)=-1.

\[Пусть\ \ t = x^{2} + 3x + 1:\]

\[t(t + 2) = - 1\]

\[t^{2} + 2t + 1 = 0\]

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

\[t + 1 = 0\]

\[t = - 1.\]

\[Подставим:\]

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

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

\[x_{1} + x_{2} = - 3;\ \ x_{1} \cdot x_{2} = 2\]

\[x_{1} = - 2;\ \ x_{2} = - 1.\]

\[Ответ:\ x = - 2;\ x = - 1.\]