Ta có:
$\frac{7x-3}{x-2}$ - $\frac{7x}{x+2}$ = $\frac{5x+4}{x^2-4}$
⇔ $\frac{(7x-3)(x+2)-7x(x-2)}{(x-2)(x+2)}$ = $\frac{5x+4}{x^2-4}$
⇔ $\frac{7x^2+14x-3x-6-7x^2+14x}{x^2-4}$ = $\frac{5x+4}{x^2-4}$
⇔ 25x - 6 = 5x + 4
⇔ 20x - 10 = 0
⇔ x = $\frac{1}{2}$
Vậy x ∈ { $\frac{1}{2}$ }
