$(x-2)^4=(x-2)^2(x-2)^2$
$=(x^2-4x+4)(x^2-4x+4)$
$=(x^2-4x+4)^2$
$=[(x^2-4x)+4]^2$
$=(x^2-4x)^2+8(x^2-4x)+16$
$=x^4-8x^3+16x^2+8x^2-32x+16$
$→x^4+(-8x^3)+24x^2+(-32x)+16=ax^4+bx^3+cx^2+dx+e$
Đồng nhất thức:
\(→\left[ \begin{array}{l}a=1\\b=-8\\c=24\\d=-32\\e=16\end{array} \right.\)
$→S=a+b+c+d+e=1-8+24-32+16=1$
Vậy $S=1$