$(x+\frac{1}{x})^2+2.(x+\frac{1}{x})-8=0$ $Đkxđ:x\neq0$
$⇔(x+\frac{1}{x})(x+\frac{1}{x}+2)-8=0$
$⇔\frac{x^2+1}{x}.\frac{x^2+2x+1}{x}-8=0$
$⇔x^4+2x^3+x^2+x^2+2x+1-8x^2=0$
$⇔x^4+2x^3-6x^2+2x+1=0$
$⇔(x^4-x^3)+(3x^3-3x^2)-(3x^2-3x)-(x-1)=0$
$⇔(x-1)(x^3+3x^2-3x-1)=0$
$⇔(x-1)(x+2-\sqrt{3})(x+2+\sqrt{3})=0$
$⇔\left[ \begin{array}{l}x=1(tm)\\x=-2+\sqrt{3}(tm)\\x=-2-\sqrt{3}(tm)\end{array} \right.$
Vậy $S=\{1;-2+\sqrt{3};-2-\sqrt{3}\}$.
