Giải thích các bước giải:

Ta có:

$\dfrac{x^3-7x-6}{x^2(x-3)^2+4x(3-x)^2+4(x-3)^2}$

$=\dfrac{x^3-x-6x-6}{x^2(x-3)^2+4x(x-3)^2+4(x-3)^2}$

$=\dfrac{x(x-1)(x+1)-6(x+1)}{(x-3)^2(x^2+4x+4)}$

$=\dfrac{(x+1)(x(x-1)-6)}{(x-3)^2(x+2)^2}$

$=\dfrac{(x+1)(x^2-x-6)}{(x-3)^2(x+2)^2}$

$=\dfrac{(x+1)(x+2)(x-3)}{(x-3)^2(x+2)^2}$

$=\dfrac{x+1}{(x-3)(x+2)}$