$x^4-4x^3+3x^2+4x-4=0$
$⇒x^4-4x^3+(4x^2-x^2)+4x-4=0$
$⇒x^4-4x^3+4x^2-x^2+4x-4=0$
$⇒(x^4-4x^3+4x^2)-(x^2-4x+4)=0$
$⇒(x^2-1)(x^2-4x+4)=0$
\(⇒\left[ \begin{array}{l}x^2-1=0⇒x^2=1⇒x=±1\\x^2-4x+4=0⇒x^2-4x=-4⇒x(x-4)=2.(-2)⇒x=2\end{array} \right.\)
vậy $x∈${$±1;2$}