a) $\frac{4x−2}{6}$ +$\frac{3−2x}{2}$ =2- $\frac{x+7}{3}$
⇔ $\frac{4x−2}{6}$ +$\frac{9−6x}{6}$ =$\frac{12}{6}$-$\frac{2x+14}{6}$
⇔ 4x-2+9-6x=12-2x-14
⇔ 4x-6x+2x=12-14+2-9
⇔ 0x=-9
⇔ x=-9
Vậy S ∈ R
b) $\frac{5}{x+2}$ - $\frac{3}{x−2}$ =$\frac{2(x−9)}{x²−4}$
ĐKXĐ:$\left \{ {{x=-2} \atop {x=2}} \right.$
⇔ $\frac{5x−10}{x²−4}$ - $\frac{3x+6}{x²−2}$ =$\frac{2x−18}{x²−4}$
⇒ 5x-10-3x+6=2x-18
⇔ 5x-3x-2x=-18+10-6
⇔ 0x=-14
Vậy S ∈ R