Đáp án:
$S=-\dfrac{x^3-x^2-3x+9}{(x-3)(2+x)}$
Giải thích các bước giải:
$36)S=\left(\dfrac{x^3-3x}{x^2-9}-1\right):\left[\dfrac{9-x^2}{(x+3)(x-2)}+\dfrac{x-3}{x-2}-\dfrac{x+2}{x+3}\right]\\ =\left(\dfrac{x^3-3x}{x^2-9}-\dfrac{x^2-9}{x^2-9}\right):\left[\dfrac{9-x^2}{(x+3)(x-2)}+\dfrac{(x-3)(x+3)}{(x-2)(x+3)}-\dfrac{(x+2)(x-2)}{(x+3)(x-2)}\right]\\ =\dfrac{x^3-x^2-3x+9}{x^2-9}:\dfrac{9-x^2+(x-3)(x+3)-(x+2)(x-2)}{(x+3)(x-2)}\\ =\dfrac{x^3-x^2-3x+9}{x^2-9}:\dfrac{−x^2+4}{(x+3)(x-2)}\\ =\dfrac{x^3-x^2-3x+9}{(x-3)(x+3)}.\dfrac{(x+3)(x-2)}{4−x^2}\\ =\dfrac{x^3-x^2-3x+9}{(x-3)(x+3)}.\dfrac{(x+3)(x-2)}{(2-x)(2+x)}\\ =-\dfrac{x^3-x^2-3x+9}{(x-3)(2+x)}$
