Giải phương trình (x^2-2x)^2 -3x^2 +6x -4 =0

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`(x^2 - 2x)^2 - 3x^2 +6x-4=0`

`⇔ (x^2 - 2x)^2 - (3x^2 - 6x)-4=0`

`⇔ (x^2 - 2x)^2 - 3 (x^2 - 2x)-4=0`

Đặt `a=x^2 - 2x`

`⇔ a^2 - 3a -4=0`

`⇔ a^2 +a - 4a - 4=0`

`⇔ (a^2 +a)-(4a+4)=0`

`⇔ a (a+1)-4(a+1)=0`

`⇔(a+1)(a-4)=0`

`⇔` \(\left[ \begin{array}{l}a+1=0\\a-4=0\end{array} \right.\) $\\$ `<=>` \(\left[ \begin{array}{l}a=-1\\a=4\end{array} \right.\) $\\$ `<=>` \(\left[ \begin{array}{l}x^2 -2x=-1\\x^2-2x=4\end{array} \right.\) $\\$ `<=>` \(\left[ \begin{array}{l}x^2-2x+1=0\\x^2-2x-4=0\end{array} \right.\) $\\$ `<=>` \(\left[ \begin{array}{l}(x-1)^2=0\\(x-1)^2=5\end{array} \right.\) $\\$ `<=>` \(\left[ \begin{array}{l}x-1=0\\x-1=\sqrt{5}\\x-1=-\sqrt{5}\end{array} \right.\) $\\$ `<=>` \(\left[ \begin{array}{l}x=1\\x=\sqrt{5}+1\\x=-\sqrt{5}+1\end{array} \right.\) 

Vậy pt có tập nghiệm : `S={1;\sqrt{5}+1; -\sqrt{5}+1}`