Question

(x^2+3x+1)(x^2+3x+3)=-1

Answer 1

(x^2+3x+1)(x^2+3x+3)=-1

Путь x^2+3x=z, тогда (z+1)(z+3)=-1

z^2+4z+4=0

D=0

z=-2

 

x^2+3x+2=0

D=1

x=-2

x=-1

 

ОТВЕТ: -2; -1

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Answer 2

$(x^2+3x+1)(x^2+3x+3)=-1 \ x^2+3x+1=t \ t(t+2)=-1 \ t^2+2t+1=0 \ (t+1)^2=0 \ t=-1 \ x^2+3x+1=-1 \ x^2+3x+2=0 \ D=b^2-4ac=(-3)^2-4*1*2=9-8=1 \ x_{1,2}=frac{-bбsqrt{D}}{2a} \ \ x_1=frac{-3+1}{2}=-1 \ \ x_2=frac{-3-1}{2}=-2$

Ответ: -1; -2