Verified answer

[tex]\displaystyle\bf\\(x^{2} +3x+1)(x^{2} +3x+3)=-1\\\\x^{2} +3x+1=m \ \ \ \Rightarrow \ \ \ x^{2} +3x+3=m+2\\\\m\cdot(m+2)=-1\\\\m^{2} +2m+1=0\\\\(m+1)^{2} =0\\\\m+1=0\\\\m=-1\\\\\\x^{2} +3x+1=-1\\\\x^{2} +3x+2=0\\\\Teorema \ Vieta:\\\\x_{1} + x_{2} =-3\\\\x_{1} \cdot x_{2} =2\\\\x_{1} =-1 \ \ ; \ \ x_{2} =-2\\\\x_{1} \cdot x_{2} =-1\cdot (-2)=2\\\\Otvet: \ 2[/tex]

Решение.

[tex]\bf (x^2+3x+1)(x^2+3x+3)=-1[/tex]

Замена переменных :  [tex]\bf t=x^2+3x+1\ \ \Rightarrow \ \ \ x^2+3x+3=t+2[/tex]  

[tex]\bf t\, (t+2)=-1\ \ ,\ \ t^2+2t+1=0\ \ ,\ \ (t+1)^2=0\ \ ,\ \ t=-1[/tex]

Перейдём к старой переменной.

[tex]\bf x^2+3x+1=-1\ \ ,\ \ \ x^2+3x+2=0\ \ ,\\\\x_1=-2\ ,\ x_2=-1\ \ (teorema\ Vieta)[/tex]  

Ответ:  [tex]\bf x_1=-2\ ,\ x_2=-1\ ,\ x_1\cdot x_2=2\ .[/tex]