Answer
$\frac{x^{2}+40x-25}{(x-4)(x+5)}; x \ne 4,-5;$
Work Step by Step
$\frac{6x^{2}+17x-40}{x^{2}+x-20} + \frac{3}{x-4} - \frac{5x}{x+5}$
Factors of $(x^{2}+x-20) $ are $(x-4)(x+5)$ So, the given expression can be written as
$= \frac{6x^{2}+17x-40}{(x-4)(x+5)} + \frac{3}{x-4} - \frac{5x}{x+5}; x \ne 4,-5;$
Taking LCD,
$= \frac{6x^{2}+17x-40 + 3(x+5) - 5x(x-4)}{(x-4)(x+5)} ; x \ne 4,-5;$
$= \frac{6x^{2}+17x-40 + 3x+15 - 5x^{2}+20x}{(x-4)(x+5)} ; x \ne 4,-5;$
Add like terms in the numerator.
$= \frac{(6-5)x^{2}+(17+3+20)x+(-40 +15)}{(x-4)(x+5)} ; x \ne 4,-5;$
$=\frac{x^{2}+40x-25}{(x-4)(x+5)}; x \ne 4,-5;$
