Ta có : \(x^3-5x^2+8x-4=0\)
=> \(x^3-x^2-4x^2+4x+4x-4=0\)
=> \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
=> \(\left(x^2-4x+4\right)\left(x-1\right)=0\)
=> \(\left(x-2\right)^2\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}\left(x-2\right)^2=0\\x-1=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy phương trình có nghiệm là \(x=1,x=2\) .