Ответы на вопрос:
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
Популярно: Алгебра
-
акл209.08.2020 07:26
-
кек78656443401.07.2022 08:57
-
Flora99925.03.2020 12:42
-
sher123452818.07.2022 00:06
-
хома7911.05.2023 15:57
-
jfjjthvhhcj29.12.2020 18:23
-
1993nnnn22.05.2023 09:25
-
kottonkandy09.03.2023 04:33
-
homya4okm03.05.2022 01:58
-
vikasivak02p08wjc18.01.2023 05:12