Answer
$\frac{-2x(x^2-2)}{(x+2)(x^2-x-3)}$
Work Step by Step
1. $\frac{x-2}{x+2}+\frac{x-1}{x+1}=\frac{(x-2)(x+1)+(x-1)(x+2)}{(x+2)(x+1)}=\frac{2x^2-4}{(x+2)(x+1)}=\frac{2(x^2-2)}{(x+2)(x+1)}$
2. $\frac{x}{x+1}-\frac{2x-3}{x}=\frac{x^2-(2x-3)(x+1)}{x(x+1)}=\frac{-x^2+x+3}{x(x+1)}$
3. Combine the above $\frac{\frac{x-2}{x+2}+\frac{x-1}{x+1}}{\frac{x}{x+1}-\frac{2x-3}{x}}=\frac{2(x^2-2)}{(x+2)(x+1)}\times\frac{x(x+1)}{-x^2+x+3}=\frac{-2x(x^2-2)}{(x+2)(x^2-x-3)}$
