Để thu gọn các biểu thức, ta sẽ thực hiện các bước sau:

**Biểu thức A:**
$A = \frac{\left(\frac{x}{x^2 - 4} + \frac{1}{x + 2} - \frac{2}{x - 2}\right)}{1 - \frac{x}{x + 2}}$
$A = \frac{\left(\frac{x}{(x - 2)(x + 2)} + \frac{1}{x + 2} - \frac{2}{x - 2}\right)}{\frac{x + 2 - x}{x + 2}}$
$A = \frac{\left(\frac{x}{(x - 2)(x + 2)} + \frac{1}{x + 2} - \frac{2}{x - 2}\right)}{\frac{2}{x + 2}}$
$A = \frac{x(x + 2) + (x - 2) - 2(x^2 - 4)}{2(x - 2)(x + 2)}$
$A = \frac{x^2 + 2x + x - 2 - 2x^2 + 8}{2(x - 2)(x + 2)}$
$A = \frac{-x^2 + 3x + 6}{2(x - 2)(x + 2)}$

**Biểu thức B:**
$B = \frac{\left(\frac{2x}{x - 3} + \frac{x}{x + 3} + \frac{2x^2 + 3x + 1}{9 - x^2}\right)}{\frac{x - 1}{x + 3}}$
$B = \frac{\left(\frac{2x}{x - 3} + \frac{x}{x + 3} + \frac{2x^2 + 3x + 1}{(3 - x)(3 + x)}\right)}{\frac{x - 1}{x + 3}}$
$B = \frac{\left(\frac{2x(x + 3) + x(x - 3) + 2x^2 + 3x + 1}{(3 - x)(3 + x)}\right)}{\frac{x - 1}{x + 3}}$
$B = \frac{\left(2x^2 + 6x + x^2 - 3x + 2x^2 + 3x + 1\right)}{(3 - x)(3 + x)} \cdot \frac{x + 3}{x - 1}$
$B = \frac{\left(5x^2 + 9x + 1\right)(x + 3)}{(3 - x)(3 + x)(x - 1)}$

**Biểu thức C:**
$C = \frac{\left(\frac{1}{x + 2} + \frac{5}{x - 2} + \frac{4}{x^2 - 4}\right)}{\frac{6}{x + 3}}$
$C = \frac{\left(\frac{1(x - 2) + 5(x + 2) + 4}{x^2 - 4}\right)}{\frac{6}{x + 3}}$
$C = \frac{\left(x - 2 + 5x + 10 + 4\right)}{(x - 2)(x + 2)} \cdot \frac{x + 3}{6}$
$C = \frac{\left(6x + 12\right)(x + 3)}{6(x - 2)(x + 2)}$

**Biểu thức D:**
$D = \frac{(x + 2)^2}{x} \cdot \frac{1 - \frac{x^2}{x + 2}}{x} - \frac{x^2 + 6x + 4}{x}$
$D = \frac{(x + 2)^2(1 - \frac{x^2}{x + 2}) - (x^2 + 6x + 4)}{x}$
$D = \frac{(x + 2)^2(x + 2 - x^2) - (x^2 + 6x + 4)}{x}$
$D = \frac{(x + 2)^2(x^2 + 2 - x^2) - (x^2 + 6x + 4)}{x}$
$D = \frac{(x + 2)^2(2) - (x^2 + 6x + 4)}{x}$
$D = \frac{2(x + 2)^2 - (x^2 + 6x + 4)}{x}$
$D = \frac{2(x^2 + 4x + 4) - x^2 - 6x - 4}{x}$
$D = \frac{2x^2 + 8x + 8 - x^2 - 6x - 4}{x}$
$D = \frac{x^2 + 2x + 4}{x}$