## Precalculus (6th Edition) Blitzer

When dividing a rational expression, we need to invert the divisor and then multiply it to the first expression $\frac{7x}{x+3}\div\frac{(x+3)^2}{x-5}=\frac{7x}{x+3}\times\frac{x-5}{(x+3)^2}=\frac{7x(x-5)}{(x+3)^3}$. Thus, the statement does not make sense.