Bài giải:

$\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x(3x-2)+1}{{{x}^{2}}-4}$

ĐKXĐ:  x $\ne $2; x $\ne $-2

$\Leftrightarrow \frac{(1-6x)(x+2)}{(x-2)(x+2)}+\frac{(9x+4)(x-2)}{(x+2)(x-2)}=\frac{x(3x-2)+1}{{{x}^{2}}-4}$

$\Leftrightarrow $ -6x² – 11x + 2 + 9x² – 14x – 8 = 3x² – 2x + 1

$\Leftrightarrow $3x² – 25x – 6 = 3x² – 2x + 1

$\Leftrightarrow $-23x = 7

$\Leftrightarrow $x = $-\frac{7}{23}$

Vậy đáp án đúng là: D