$x^{2}$ (x-2)+16(2-x)=0
<=> $x^{2}$ (x-2) - 16 ( x-2) =0
<=> (x-2) ($x^{2}$ -16)
<=> \(\left[ \begin{array}{l}x-2=0\\x^{2} -16 = 0\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=2\\x^{2} = 16\end{array} \right.\)
<=> \(\left[ \begin{array}{l}x=2\\x= ± 4\end{array} \right.\)
Vậy S = {2; ±4}
#Trang Huyen