`\frac{x^2-4x+7}{x+1}-2=-\frac{x^2-3x+7}{2x+1}` (Điều kiện: `x \ne -1; x \ne -\frac{1}{2}`)
`⇔\frac{(x^2-4x+7)(2x+1)}{(x+1)(2x+1)}=-\frac{2(x+1)(x^2-3x+7)}{(x+1)(2x+1)}`
`⇒2x^3-7x^2+10x+7-4x^2-6x-2+x^3-2x^2+4x+7=0`
`⇔3x^3-13x^2+8x+12=0`
`⇔3x^3-4x^2-4x-9x^2+12x+12=0`
`⇔x(3x^2-4x-4)-3(3x^2-4x-4)=0`
`⇔(x-3)(3x^2-4x-4)=0`
`⇔(x-3)(3x^2-6x+2x-4)=0`
`⇔(x-3)[3x(x-2)+2(x-2)]=0`
`⇔(x-3)(x-2)(3x+2)=0`
`⇔`\(\left[ \begin{array}{l}x-3=0\\x-2=0\\3x+2=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3\\x=2\\3x=-2\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=3(\text{thỏa mãn điều kiện})\\x=2(\text{thỏa mãn điều kiện})\\x=-\dfrac{2}{3}(\text{thỏa mãn điều kiện})\end{array} \right.\)
Vậy: `S={3;2;-\frac{2}{3}}`
