x^4+x²+6x-8=0
⇔(x^4-x³)+(x³-x²)+(2x²-2x)+(8x-8)=0
⇔(x-1)(x³+x²+2x+8)=0
⇔(x-1)[(x³+2x²)-(x²+2x)+(4x+8)]=0
⇔(x-1)(x+2)(x²-x+4)=0
⇔(x-1)(x+2)[(x-$\frac{1}{2}$)²+$\frac{15}{4}$]=0
Vì (x-$\frac{1}{2}$)²≥0∀x ⇒ (x-$\frac{1}{2}$)²+$\frac{15}{4}$>0∀x
⇒(x-1)(x+2)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x+2=0\end{array} \right.\)⇔\(\left[ \begin{array}{l}x=1\\x=-2\end{array} \right.\)
Vậy S={1;-2}.
