(x^2-3x+1)^2+3(x-1) (x^2-3x+1)=4(x-1)^2 молю - вопрос №3071376231231231412 от vevvrqc 04.06.2023 11:01

Question

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

vevvrqc · Answer

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)]=aa²+3a-4=0a1+a2=-3 U a1*a2=-4a1=-4⇒[(x²-3x+1)/(x-1)]=-4[(x²-3x+1)/(x-1)]+4=0x²-3x+1+4x-4=0x²+x-3=0D=1+12=13x1=(-1-√13)/2x2=(-1+√13)/2a2=1⇒[(x²-3x+1)/(x-1)]=1[(x²-3x+1)/(x-1)]-1=0x²-3x+1-x+1=0x²-4x+2=0D=16-8=8x3=(4-2√2)/2=2-√2x2=2+√2...