Đáp án:
$S=\{\sqrt{5}-2;-\sqrt 5+2;\sqrt 5+2;-\sqrt 5-2\}$
Giải thích các bước giải:
$x^4-18x^2+1=0$
Đặt $t=x^2\ge 0$
$⇔t^2-18t+1=0$
$Δ'=9^2-1.1=80⇒\sqrt{Δ'}=4\sqrt{5}$
$⇒\left[ \begin{array}{l}t=9-4\sqrt{5}=5-4\sqrt{5}+4=(\sqrt{5}-2)^2\\t=9+4\sqrt{5}=5+4\sqrt{5}+4=(\sqrt{5}+2)^2\end{array} \right.$
$⇒\left[ \begin{array}{l}x^2=(\sqrt{5}-2)^2\\x^2=(\sqrt{5}+2)^2\end{array} \right.⇒\left[ \begin{array}{l}x=|\sqrt{5}-2|\\x=-|\sqrt{5}-2|\\x=|\sqrt{5}+2|\\x=-|\sqrt{5}+2|\end{array} \right.⇒\left[ \begin{array}{l}x=\sqrt{5}-2\\x=-\sqrt{5}+2\\x=\sqrt{5}+2\\x=-\sqrt{5}-2\end{array} \right.$
Vậy $S=\{\sqrt{5}-2;-\sqrt 5+2;\sqrt 5+2;-\sqrt 5-2\}$.
