Đáp án: $S=\{±\sqrt{3};\dfrac{-1}{2};1\}$
Giải thích các bước giải:
$2x^4-x^3-7x^2+3x+3=0$
$⇔2x^4-2x^3+x^3-x^2-6x^2+6x-3x+3=0$
$⇔2x^3(x-1)+x^2(x-1)-6x(x-1)-3(x-1)=0$
$⇔(2x^3+x^2-6x-3)(x-1)=0$
$⇔[x^2(2x+1)-3(2x+1)](x-1)=0$
$⇔(x^2-3)(2x+1)(x-1)=0$
$⇔(x-\sqrt{3})(x+\sqrt{3})(2x+1)(x-1)=0$
$⇔\left[ \begin{array}{l}x-\sqrt{3}=0\\x+\sqrt{3}=0\\2x+1=0\\x-1=0\end{array} \right.⇔\left[ \begin{array}{l}x=\sqrt{3}\\x=-\sqrt{3}\\x=\dfrac{-1}{2}\\x=1\end{array} \right.$
