Bài giải:

$\frac{{{x}^{3}}-{{(x-1)}^{3}}}{(4x+3)(x-5)}=\frac{7x-1}{4x+3}-\frac{x}{x-5}$

ĐKXĐ:  x $\ne $$-\frac{3}{4}$ ; x $\ne $5

$\Leftrightarrow \frac{{{x}^{3}}-{{x}^{3}}+3{{x}^{2}}-3x+1}{(4x+3)(x-5)}=\frac{(7x-1)(x-5)-x(4x+3)}{(4x+3)(x-5)}$

$\Leftrightarrow $3x² – 3x + 1 = 7x² – 36x + 5 – 4x² – 3x

$\Leftrightarrow $3x² – 3x + 1 = 3x² – 39x + 5

$\Leftrightarrow $36x = 4

$\Leftrightarrow $x = $\frac{1}{9}$

Vậy đáp án đúng là: A