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