Đáp án:

$x^{4}$ $-$ $2x^{2}$ =$8$ $-$ $4x^{2}$ 
$<=>$$x^{4}$ $-$ $2x^{2}$ $-$ $8$ $+$ $4x^{2}$ $=$ $0$
$<=>$$x^{4}$+ $2x^{2}$ $-$ $8$ $=$ $0$ $(1)$
$Thay$ $y$ $=$ $x^{2}$ $vào$ $pt$$,$ $ta$ $được$$:$
$y^{2}$$+$ $2y$ $-$ $8$$=0$
$<=>$$(y+4)$$.$$(y-2)$$=$$0$
$<=>$ \(\left[ \begin{array}{l}y+4=0\\y-2=0\end{array} \right.\) 
$<=>$\(\left[ \begin{array}{l}y=-4\\y=2\end{array} \right.\) 
$Thay$ $x^{2}$ $=$ $-4$ $vào$ $(1)$$,$ $ta$ $được$$:$
$-4x^{2}$ $-2.(-4)$ $-8=0$
$<=>$ $-4x^{2}$ $+8-8=0$
$<=>$$-4x^{2}$ $=0$
$<=>$$x^{2}$ $=0$
$<=>$$x=0$
$Thay$ $x^{2}=2$  $vào$ $(1)$$,$$ta$ $được$$:$
$2x^{2}$ $+2.2-8=0$
$<=>$$2x^{2}$ $-4=0$
$<=>$ $2.$$(x^{2} -2)$$=0$
$<=>$$x^{2}$ $-2=0$
$<=>$$x^{2}$ $=2$
$<=>$$x=√2$
$Vậy$ $pt$ $có$ $nghiệm$$:$ $S$$=${$√2;0$}
$Chúc$ $bạn$ $học$ $tốt$$!$
$Answered$ $by$ $Mai$ $Chi$$.$

x^4-x^2+4x^2-8=0

x^4+2x^2+1+9=0

(x^2+1)^2+3^2=0

(x^2+1)^2=-3^2=3^2

x^2+1=3

x^2=2

x=√2