$(x^2+2x)^2-x^2-2x-2=0$ $(1)$
$(x^2+2x)^2-(x^2+2x+2)=0$
Đặt $x^2+2x=t$
Từ $(1)⇒t^2-t-2=0$
$⇔(t^2-2t)+(t-2)=0$
$⇔(t-2)(t+1)=0$
$⇔\left[ \begin{array}{l}t-2=0\\t+1=0\end{array} \right.⇔\left[ \begin{array}{l}t=2\\t=-1\end{array} \right.$
Với $t=2⇒x^2+2x-2=0$
$⇔(x^2+2x+1)-3=0$
$⇔(x+1)^2-3=0$
$⇔(x+1+\sqrt{3})(x+1-\sqrt{3})=0$
$⇔\left[ \begin{array}{l}x=-1-\sqrt{3}\\x=-1+\sqrt{3}\end{array} \right.$
Với $t=-1⇒x^2+2x+1=0$
$⇔(x+1)^2=0$
$⇔x=-1$
Vậy $S=\{-1-\sqrt{3};-1+\sqrt{3};-1\}$
$x^4-x^3-x^2-x-2=0$
$⇔(x^4-2x^3)+(x^3-2x^2)+(x^2-2x)+(x-2)=0$
$⇔(x-2)(x^3+x^2+x+1)=0$
$⇔(x-2)[(x^3+x^2)+(x+1)]=0$
$⇔(x-2)(x+1)(x^2+1)=0$
Vì $x^2≥0∀x⇒x^2+1>0∀x$
$⇒(x-2)(x+1)=0$
$⇔\left[ \begin{array}{l}x-2=0\\x+1=0\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=-1\end{array} \right.$
Vậy $S=\{2;-1\}$
$x^3-2x^2-9x+18=0$
$⇔(x^3-2x^2)-(9x-18)=0$
$⇔(x-2)(x^2-9)=0$
$⇔(x-2)(x+3)(x-3)=0$
$⇔\left[ \begin{array}{l}x-2=0\\x-3=0\\x+3=0\end{array} \right.⇔\left[ \begin{array}{l}x=2\\x=3\\x=-3\end{array} \right.$
Vậy $S=\{2;3;-3\}$.
