Chapter 6 - Section 6.2 - Adding and Subtracting Rational Expressions - Exercise Set - Page 427: 42

$\frac{2}{x+5}$

$x^2+3x-10=x^2+5x-2x-10=x(x+5)-2(x+5)=(x+5)(x-2)$, so LCD$=(x+5)(x-2)$, thus $\frac{x^2-39}{x^2+3x-10}-\frac{x-7}{x-2}=\frac{x^2-39}{(x+5)(x-2)}-\frac{(x-7)(x+5)}{(x+5)(x-2)}=\frac{x^2-39-(x^2-2x-35)}{(x+5)(x-2)}=\frac{2x-4}{(x+5)(x-2)}=\frac{2(x-2)}{(x+5)(x-2)}=\frac{2}{x+5}$