(x² - x + 1)² -3 (x² - x + 1) -4 =0
Đặt t=x²- x + 1; ta có
(x²- x + 1)²-3 (x² - x + 1) -4=0
(=)t²-3t-4=0
(=)t²+t-4t-4=0
(=)t(t+1)-4(t+1)=0
(=)(t-4)(t+1)=0
Thay t=x²- x + 1; ta có
(t-4)(t+1)=0
(=)(x²- x + 1-4)(x²- x + 1+1)=0
(=)(x²-x-3)(x²-x+2)=0
(=)(x²-x-3)(x²+x-2x+2)=0
(=)(x²-x-3)[x(x+1)-2(x+1)]=0
(=)(x²-x-3)(x-2)(x+1)=0
1)x-2=0=)x=2
2)x+1=0=)x=-1
3)x²-x-3=0=)
- x =(1-√13)/2=-1.303
- x =(1+√13)/2= 2.303
