(x^2-3x+1)^2+3(x-1) (x^2-3x+1)=4(x-1)^2 молю о !

Разделим каждое на (х-1)²≠0
[(x²-3x+1)/(x-1)]²+3[(x²-3x+1)/(x-1)]-4=0
[(x²-3x+1)/(x-1)]=a
a²+3a-4=0
a1+a2=-3 U a1*a2=-4
a1=-4⇒[(x²-3x+1)/(x-1)]=-4
[(x²-3x+1)/(x-1)]+4=0
x²-3x+1+4x-4=0
x²+x-3=0
D=1+12=13
x1=(-1-√13)/2
x2=(-1+√13)/2
a2=1⇒[(x²-3x+1)/(x-1)]=1
[(x²-3x+1)/(x-1)]-1=0
x²-3x+1-x+1=0
x²-4x+2=0
D=16-8=8
x3=(4-2√2)/2=2-√2
x2=2+√2