Bài giải:

M = $1-\left( \frac{16x-{{x}^{2}}}{{{x}^{2}}-4}+\frac{3+2x}{x-2}+\frac{3x-2}{x+2} \right):\frac{x-1}{{{x}^{2}}+4x+4}$

M = 1 - $\left( \frac{16x-{{x}^{2}}}{(x-2)(x+2)}+\frac{3+2x}{x-2}+\frac{3x-2}{x+2} \right):\frac{x-1}{{{(x+2)}^{2}}}$

M = $1-\frac{16x-{{x}^{2}}+(3+2x)(x+2)+(3x-2)(x-2)}{(x-2)(x+2)}:\frac{x-1}{{{(x+2)}^{2}}}$

M = $1-\frac{16x-{{x}^{2}}-2{{x}^{2}}-7x-6+3{{x}^{2}}-8x+4}{(x-2)(x+2)}.\frac{{{(x+2)}^{2}}}{x-1}$

M = $1-\frac{x-2}{(x-2)(x+2)}.\frac{{{(x+2)}^{2}}}{x-1}$

M = $1-\frac{x+2}{x-1}$ = $\frac{x-1-x-2}{x-1}$ = $\frac{-3}{x-1}$

Vậy đáp án đúng là: D