Giải thích các bước giải:
$\dfrac{2}{(x-3)(x-1)}+\dfrac{-7}{(x+4)(x-3)}+\dfrac{4}{x(x+4)}$
$=\dfrac{2x(x+4)}{x(x+4)(x-3)(x-1)}+\dfrac{-7x(x-1)}{x(x-1)(x+4)(x-3)}+\dfrac{4(x-1)(x-3)}{(x-1)(x-3)x(x+4)}$
$=\dfrac{2x^2+8x}{x(x+4)(x-3)(x-1)}+\dfrac{-7x^2+7x}{x(x-1)(x+4)(x-3)}+\dfrac{4(x^2-4x+3)}{(x-1)(x-3)x(x+4)}$
$=\dfrac{2x^2+8x-7x^2+7x+4(x^2-4x+3)}{(x-1)(x-3)x(x+4)}$
$=\dfrac{-x^2-x+12}{(x-1)(x-3)x(x+4)}$
$=\dfrac{-(x^2+4x-3x-12)}{(x-1)(x-3)x(x+4)}$
$=\dfrac{-(x+4)(x-3)}{(x-1)(x-3)x(x+4)}$
$=\dfrac{-}{(x-1)x}$
