## College Algebra (10th Edition)

$-\displaystyle \frac{2x(x^{2}-2)}{(x+2)(x^{2}-x-3)}$
$\displaystyle \frac{\frac{x-2}{x+2}+\frac{x-1}{x+1}}{\frac{x}{x+1}-\frac{2x-3}{x}}=(\frac{x-2}{x+2}+\frac{x-1}{x+1})\div(\frac{x}{x+1}-\frac{2x-3}{x})$ 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... $=-\displaystyle \frac{2x(x^{2}-2)}{(x+2)(x^{2}-x-3)}$