Đáp án:
Giải thích các bước giải:
2) `(x^2-2x)^2+|x^2-2x|-2=0`
Đặt `x^2-2x=t`, ta có:
`t^2+|t|-2=0`
`⇔ |t|=2-t^2`
TH1: `t \ge 0⇔ x^2-2x \ge 0 ⇔` \(\left[ \begin{array}{l}x \le 0\\x \ge 2\end{array} \right.\)
`⇔ t=2-t^2`
`⇔ t^2+t-2=0`
`⇔ (t-1)(t+2)=0`
`⇔` \(\left[ \begin{array}{l}t=-2\\t=1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x^2-2x=-2\\x^2-2x=1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x^2-2x+2=0\ \text{(vô nghiệm do }(x-1)^2+1 \ge 1)\\x^2-2x-1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1+\sqrt{2}\\x=1-\sqrt{2}\end{array} \right.\) (TM)
TH2: `t < 0⇔ x^2-2x < 0 ⇔ 0<x<2`
`t=t^2-2`
`⇔ t^2-t-2=0`
`⇔ (t+1)(t-2)=0`
`⇔` \(\left[ \begin{array}{l}t=2\\t=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x^2-2x=2\\x^2-2x=-1\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x^2-2x-2=0\\x^2-2x+1=0\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=1+\sqrt{3}\ (L)\\x=1-\sqrt{3}\ (L)\\x=1\ (TM)\end{array} \right.\)
Vậy `S={1+\sqrt{2},1-\sqrt{2},1}`
