Giải thích các bước giải:
$A=(\dfrac{3+x}{3-x}-\dfrac{3-x}{x+3}+\dfrac{4x^2}{x^2-9}).(2+x+\dfrac{10-x^2}{x+2})$
$\to A=(\dfrac{(3+x)^2-(3-x)^2}{(3-x)(x+3)}+\dfrac{4x^2}{(x-3)(x+3)}).\dfrac{(x+2)^2+10-x^2}{x+2}$
$\to A=\dfrac{(3+x)^2-(3-x)^2-4x^2}{(3-x)(x+3)}.\dfrac{(x+2)^2+10-x^2}{x+2}$
$\to A=\dfrac{-4x^2+12x}{(3-x)(x+3)}.\dfrac{4x+14}{x+2}$
$\to A=\dfrac{4x(3-x)}{(3-x)(x+3)}.\dfrac{4x+14}{x+2}$
$\to A=\dfrac{4x}{x+3}.\dfrac{4x+14}{x+2}$
