Giải thích các bước giải:
$\dfrac{x}{x+1}+\dfrac{3-3x}{x^2-x+1}-\dfrac{x+4}{x^3+1}$
$=\dfrac{x(x^2-x+1)}{(x+1)(x^2-x+1)}+\dfrac{(3-3x)(x+1)}{(x+1)(x^2-x+1)}-\dfrac{x+4}{(x+1)(x^2-x+1)}$
$=\dfrac{x(x^2-x+1)+(3-3x)(x+1)-(x+4)}{(x+1)(x^2-x+1)}$
$=\dfrac{x^3-4x^2-1}{(x+1)(x^2-x+1)}$