Answer
$-\displaystyle \frac{2x(x^{2}-2)}{(x+2)(x^{2}-x-3)}$
Work Step by Step
$\displaystyle \frac{\frac{x-2}{x+2}+\frac{x-1}{x+1}}{\frac{x}{x+1}-\frac{2x-3}{x}}=(\frac{x-2}{x+2}+\frac{x-1}{x+1})\div(\frac{x}{x+1}-\frac{2x-3}{x})$
Simplify separately the dividend and the divisor
$\displaystyle \frac{x-2}{x+2}+\frac{x-1}{x+1}=\frac{x-2}{x+2}\cdot\frac{x+1}{x+1}+\frac{x-1}{x+1}\frac{x+2}{x+2}$
$=\displaystyle \frac{x^{2}-x-2+x^{2}+x-2}{(x+2)(x+1)}$
$=\displaystyle \frac{2x^{2}-4}{(x+2)(x+1)}$
$=\displaystyle \frac{2(x^{2}-2)}{(x+2)(x+1)}$
$(\displaystyle \frac{x}{x+1}-\frac{2x-3}{x})=\frac{x}{x+1}\cdot\frac{x}{x}-\frac{2x-3}{x}\cdot\frac{x+1}{x+1}$
$=\displaystyle \frac{x^{2}-(2x^{2}+2x-3x-3)}{x(x+1)}$
$=\displaystyle \frac{-x^{2}+x+3}{x(x+1)}=\frac{-(x^{2}-x-3)}{x(x+1)}$
... no two factors of -3 add up to -1, so the numerator remains as is.
Now,
$\displaystyle \frac{\frac{x-2}{x+2}+\frac{x-1}{x+1}}{\frac{x}{x+1}-\frac{2x-3}{x}}=\frac{2(x^{2}-2)}{(x+2)(x+1)}\div\frac{-(x^{2}-x-3)}{x(x+1)}$
... division = multiplication with the reciprocal,
$=\displaystyle \frac{2(x^{2}-2)}{(x+2)(x+1)}\cdot\frac{-x(x+1)}{(x^{2}-x-3)}$
... cancel: $(x+1)$, there is one minus...
$=-\displaystyle \frac{2x(x^{2}-2)}{(x+2)(x^{2}-x-3)}$
