Đáp án:
$85 ) \dfrac{7x-3}{x-2} - \dfrac{7x}{x+2} = \dfrac{5x+4}{x^2-4}$
$\text{ĐKXĐ : x $\neq$ ±2}$
$⇔ \dfrac{(7x-3)(x+2)}{(x-2)(x+2)} - \dfrac{7x(x-2)}{(x+2)(x-2)} = \dfrac{5x+4}{(x-2)(x+2)}$
$⇒ (7x-3)(x+2) - 7x(x-2) = 5x+4$
$⇔ 7x^2 +14x -3x -6 -7x^2 +14x = 5x+4$
$⇔ 7x^2 -7x^2 +14x -3x-5x +14x = 4 +6$
$⇔ 20x = 10$
$⇔x = 10 : 20$
$⇔x = \dfrac{1}{2}$(TM)
$\text{Vậy phương trình có tập nghiệm S={$\dfrac{1}{2}$}}$
