Answer
$\frac{7(x+1)^2}{2x^2}$, $x\ne -1, -5, 0, 1, 5$
Work Step by Step
Factor each numerator and denominator and cancel common factors, provided $x\ne -1, -5, 0, 1, 5$
$\frac{x(x+5)(x-5)}{4x^2}\cdot \frac{2(x+1)(x-1)}{(x-1)(x-5)}\div \frac{x(x+5)}{7(x+1)}=\frac{(x+5)}{4x}\cdot \frac{2(x+1)}{1}\times \frac{7(x+1)}{x(x+5)}=\frac{7(x+1)^2}{2x^2}$
