jmc

First, we factor the denominators, to get $$\frac{G}{x+5}+\frac{H}{x(x-4)}=\frac{x^2-2x+10}{x(x+5)(x-4)}.$$We then multiply both sides by $x(x+5)(x-4)$, to get $$Gx(x-4)+H(x+5)=x^2-2x+10.$$We can solve for $G$ and $H$ by substituting suitable values of $x$. For example, setting $x=-5$, we get $45G=45$, so $G=1$. Setting $x=0$, we get $5H=10$, so $H=2$. (This may not seem legitimate, because we are told that the given equation holds for all $x$ except $-5$, 0, and 4. This tells us that the equation $Gx(x-4)+H(x+5)=x^2-2x+10$ holds for all $x$, except possibly $-5$, 0, and 4. However, both sides of this equation are polynomials, and if two polynomials are equal for an infinite number of values of $x$, then the two polynomials are equal for all values of $x$. Hence, we can substitute any value we wish to into this equation.)

Therefore, $H/G=2/1=\boxed{2}$.