ĐK: [TEX]x \in \mathbb{R}[/TEX]
[TEX]PT\Leftrightarrow 2(x^2-2x+2)+3(x^2+2x+2)=7\sqrt{(x^2-2x+2)(x^2+2x+2)}[/TEX]
Đặt [TEX]a=\sqrt{x^2-2x+2}; b=\sqrt{x^2+2x+2}(a;b>0) \Rightarrow 2a^2-7ab+3b^2=0[/TEX]
[TEX]\Leftrightarrow (a-3b)(2a-b)=0\Rightarrow \begin{bmatrix} a=3b & \\ 2a=b & \end{bmatrix}[/TEX]
TH1: [TEX]a=3b \Rightarrow x^2-2x+2=9(x^2+2x+2) \Leftrightarrow 8x^2+20x+16=0[/TEX] (vô nghiệm vì [TEX]\Delta =-112<0[/TEX])
TH2:[TEX]2a=b\Leftrightarrow 4(x^2-2x+2)=(x^2+2x+2)\Leftrightarrow 3x^2-10x+6[/TEX] [TEX]\Rightarrow x=\frac{5\pm \sqrt{7}}{3}[/TEX] (thoả mãn)
Vậy [TEX]S=\left \{ \frac{5\pm \sqrt{7}}{3} \right \}[/TEX]