Đáp án:

Giải thích các bước giải:

`(x^2-2x)^2+|x^2-2x|-2=0`

`<=>|x^2-2x|^2+|x^2-2x|-2=0`

Đặt `a=|x^2-2x| (a>=0)`

`=>a^2+a-2=0`

`<=>(a+2)(a-1)=0`

Lại có `a>=0=>a+2>=2>0`

`=>a-1=0`

`<=>|x^2-2x|-1=0`

`<=>|x^2-2x|=1`

`<=>`\(\left[ \begin{array}{l}x^2-2x=1\\x^2-2x=-1\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}x^2-2x-1=0\\x^2-2x+1=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}(x-1)^2=2\\(x-1)^2=0\end{array} \right.\)

`<=>`\(\left[ \begin{array}{l}\left[ \begin{array}{l}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{array} \right.\\x=1\end{array} \right.\)

Vậy `S={1,1+-\sqrt{2}}`

`(x^2-2x)^2+|x^2-2x|-2=0`

`<=> |x^2-2x|^2+|x^2-2x|-2=0`

Đặt `|x^2-2x|=t \ \ (t>=0`

Phương trình trở thành: `t^2+t-2=0`

`<=> t^2-t+2t-2=0`

`<=> t.(t-1)+2.(t-1)=0`

`<=> (t+2).(t-1)=0`

`<=>` \(\left[ \begin{array}{l}t=-2 \ \ \ \rm (ktm)\\t=1 \ \ \ \rm (tm)\end{array} \right.\) 

`t=1 <=> |x^2-2x|=1`

`<=>` \(\left[ \begin{array}{l}x^2-2x=1\\x^2-2x=-1\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}x^2-2x-1=0\\x^2-2x+1=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}(x-1)^2=2\\(x-1)^2=0\end{array} \right.\) `<=>` \(\left[ \begin{array}{l}\left[ \begin{array}{l}x-1=\sqrt{2}\\x-1=-\sqrt{2}\end{array} \right.\\x=1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}\left[ \begin{array}{l}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{array} \right.\\x=1\end{array} \right.\) 

Vậy `S={1+-sqrt(2);1}`