Đáp án+Giải thích các bước giải:
a,
$x^4-4x^3+8x+3=0$
$⇔x^4+x^3-5x^3-5x^2+5x^2+5x+3x+3=0$
$⇔x^3(x+1)-5x^2(x+1)+5x(x+1)+3(x+1)=0$
$⇔(x+1)(x^3-5x^2+5x+3)=0$
$⇔(x+1)(x^3-3x^2-2x^2+6x-x+3)=0$
$⇔(x+1)[x^2(x-3)-2x(x-3)-(x-3)]=0$
$⇔(x+1)(x^2-2x-1)(x-3)=0$
$⇔(x+1)(x-3)(x^2-2x+1-2)=0$
$⇔(x+1)(x-3)[(x-1)^2-2]=0$
$⇔(x+1)(x-3)(x-1+\sqrt{2})(x-1-\sqrt{2})=0$
$⇔\left[\begin{matrix}x+1=0\\x-3=0\\x-1+\sqrt{2}=0\\x-1-\sqrt{2}=0\end{matrix}\right.$
$⇔\left[\begin{matrix}x=-1\\x=3\\x=1-\sqrt{2}\\x=1+\sqrt{2}\end{matrix}\right.$
Vậy `S={-1;3;1-\sqrt{2};1+\sqrt{2}}`
