Answer
$\dfrac{x-3}{x+7}
$
Work Step by Step
By factoring both the denominator and numerator of each rational function and cancelling the common factors, we have
$$
\frac{x^2-3x-10}{x^2+2x-35}\frac{x^2+4x-21}{x^2+9x+14}=\frac{(x-5)(x+2)}{(x-5)(x+7)}\frac{(x-3)(x+7)}{(x+2)(x+7)}\\
=\frac{x-3}{x+7}
.$$
