\(x^4-x^2+2x-1=x^4-\left(x^2-2x+1\right)=x^4-\left(x-1\right)^2=\left(x^2-x+1\right)\left(x^2+x-1\right)=0\) vì \(x^2+x+1>0\) với mọi x
Do đó: \(x^2+x-1=0\Leftrightarrow x=\dfrac{-1\pm\sqrt{5}}{2}\)
Vậy x = \(\dfrac{-1+\sqrt{5}}{2};x=\dfrac{-1-\sqrt{5}}{2}\)