(х+1)^2 (х^2+2х)=0 (Х^2+3х+1)(х^2+3х+3)=-1 (Х^2-4х+1)(х^2-4х+2)=12 Решите пжж

(х+1)^2 (х^2+2х)=0 x+1=0 x=-1 x²+2x=0 x(x+2)=0 x=-2 x=0 (Х^2+3х+1)(х^2+3х+3)=-1 (x²+3x+1)(x²+3x+3)+1=0 (t+1)(t+3)+1=0 (t²+3t+t+3)+1=0 (t²+4t+3)+1=0 t²+4t+3+1=0 t²+4t+4=0 D=b²-4ac=4²-4*1*4=0 t₁.₂=-b/2a=-4/2*1=-2 x²+3x=-2 x²+3x+2=0 X₁.₂-b+-√D /2a x₁=-3-1/2*1=-2 x₂=-3+1/2*1=-1 x=-2 x=-1 (Х^2-4х+1)(х^2-4х+2)=12 (t+1)(t+2)-12=0 (t²+2t+t+2)-12=0 (t²+3t+2)-12=0 t²+3t+2-12=0 D=b²-4ac=3²-4*1(-10)=49 t₁.₂=-b+-√D/2a t₁=-3-7/2*1=-5 t₂=-3+7/2*1=2 x²-4x=-5 x²-4x=2 x²-4x=-5 x²-4x+5=0 x²-4x=2 x²-4x-2=0 D=b²-4ac=(-4)²=4*1(-2)=24 x₁.₂=-b+-√D/2a x₁=4-2√6/2*1=2-√6 x₂=4+2√6/2*1=2+√6 x=2-√6 x=2+√6 ответ:x=2-√6 x=2+√6 (х^2+1)(х^2+2х)=12 (x²+1)(x²+2x)-12=0 (x^4+2x³+x²+2x)-12=0 x^4+2x³+x²+2x-12=0