\(x^3-4x^2+2x^2-1\)
= \(x^3-2x^2+2x-1\)
= \(x^3-x^2-x^2+x+x-1\)
= \(x^2\left(x-1\right)-x\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2-x+1\right)\)
Ta thấy \(x^2-x+1=\left(x-\dfrac{1}{4}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Vậy x=1