Giải thích các bước giải:
$A=\dfrac{3x^3-2x^2}{(3x-2)(3x+2)}$
$=\dfrac{x^2(3x-2)}{(3x-2)(3x+2)}$
$=\dfrac{x^2}{3x+2}$
$B=\dfrac{x^3+27}{2x^2-6x+18}$
$=\dfrac{(x+3)(x^2-3x+9)}{2(x^2-3x+9)}$
$=\dfrac{x+3}{2}$
$C=\dfrac{x(3-x)}{x^2-9}$
$=-\dfrac{x(x-3)}{(x+3)(x-3)}$
$=-\dfrac{x}{x+3}$
Học tốt!!!