<=> (x2 - 2x)2 + x2 - 2x + 1 - 13 = 0
<=> (x2 - 2x)2 + x2 - 2x - 12 = 0
Đặt t = x2 - 2x
Khi đó ta có pt: t2 + t - 12 = 0
<=> t2 + 4t - 3t - 12 = 0
<=> (t - 3)(t + 4) = 0 <=> \(\orbr{\begin{cases}t=3\\t=-4\end{cases}}\)
*Với t = 3 ta có: x2 - 2x = 3
<=> x2 - 2x - 3 = 0
<=> (x - 3)(x + 1) = 0 <=> \(\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
*Với t = -4 ta có: x2 - 2x = -4
<=> x2 - 2x + 4 = 0
<=> (x - 1)2 + 3 = 0 (Vô nghiệm)
Vậy S = {3;-1}
(x2-2x)2 + (x-1)2 - 13 = 0
<=> x^4 - 4x^3 + 4x^2 + x^2 - 2x + 1 - 13 = 0
<=> x^3 - 4x^3 + 5x^2 - 2x - 12 = 0
<=> x^4 + x^3 - 5x^3 - 5x^2 + 10x^2 + 10x - 12x - 12 = 0
<=> x^3(x + 1) - 5x^2(x + 1) + 10x(x + 1) - 12(x + 1) = 0
<=> (x^3 - 5x^2 + 10x - 12)(x + 1) = 0
<=> (x^3 - 3x^2 - 2x^2 + 6x + 4x - 12)(x + 1) = 0
<=> [x^2(x - 3) - 2x(x - 3) + 4(x - 3)](x + 1) = 0
<=> (x^2 - 2x + 4)(x - 3)(x + 1) = 0
có x^2 - 2x + 4 = (x - 1)^2 + 3 lớn hơn 0
<=> x - 3 = 0 hoặc x + 1 = 0
<=> x = 3 hoặc x = -1