Đáp án:
$x^4 -3x^3 +x^2-9x-6=0$
$⇔x^-3x^3-2x^2+3x^2-9x-6=0$
$⇔(x^4+3x^2)+(-3x^3-9x)+(-2x^2-6)=0$
$⇔x^2(x^2+3)-3x(x^2+3)-2(x^2+3)=0$
$⇔(x^2+3).(x^2-3x-2)=0$
Vì $x^2 ≥ 0 ⇔ x^2 +3 > 0$ (loại)
$⇔x^2-3x-2=0$
$⇔x^2-2.x . \dfrac{3}{2} + \dfrac{9}{4} - \dfrac{17}{4} =0$
$⇔(x-\dfrac{3}{2})^2 - \dfrac{17}{4} =0$
$⇔(x-\dfrac{3}{2} - \sqrt[]{\dfrac{17}{4}}).(x-\dfrac{3}{2} + \sqrt[]{\dfrac{17}{4}})=0$
$⇔$\(\left[ \begin{array}{l}x-\dfrac{3}{2} - \sqrt[]{\dfrac{17}{4}}=0\\x-\dfrac{3}{2} + \sqrt[]{\dfrac{17}{4}}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=\dfrac{3+\sqrt[]{17}}{2}\\x=\dfrac{3-\sqrt[]{17}}{2}\end{array} \right.\)
Vậy S = {$\dfrac{3+\sqrt[]{17}}{2} ; \dfrac{3-\sqrt[]{17}}{2}$}
