X³-x²-8/(x³-x²)=2 ОДЗ: x³-x²≠0 x²*(x-1)≠0 x₁≠0 x₂≠1
Пусть x³-x²=t ⇒
t-8/t=2
t²-8=2t
t²-2t-8=0 D=36
t₁=4
x³-x²=4
x³-2x²+x²-4=0
x²*(x-2)+(x-2)*(x+2)=0
(x-2)*(x²+x+2)=0
x-2=0
x₁=2
x²+x+2=0 D=-7∉
t₂=-2
x³-x²=-2
x³-x²+2=0
x³+x²-x²-x²+2=0
x³+x²-(2x²-2)=0
x²*(x+1)-2*(x²-1)=0
x²*(x+1)-2*(x+1)*(x-1)=0
(x+1)*(x²-2x+2)=0
x+1=0
x₂=-1
x²-2x+2=0 D=-4∉
Ответ: x₁=2 x₂=-1.