Giải thích các bước giải:
a.Ta có :
$\dfrac{3x}{2x+4}+\dfrac{x+3}{x^2-4}$
$=\dfrac{3x(x-2)}{2(x+2)(x-2)}+\dfrac{2(x+3)}{2(x+2)(x-2)}$
$=\dfrac{3x(x-2)+2(x+3)}{2(x+2)(x-2)}$
$=\dfrac{3x^2-4x+6}{2(x+2)(x-2)}$
b.Ta có :
$\dfrac{2x}{x+2}-\dfrac{x+3}{x^2-4}$
$=\dfrac{2x(x-2)}{(x+2)(x-2)}-\dfrac{x+3}{(x+2)(x-2)}$
$=\dfrac{2x(x-2)-x-3}{(x+2)(x-2)}$
$=\dfrac{2x^2-5x-3}{(x+2)(x-2)}$