Answer
$\frac{x+7}{x}$ and $x\ne7, x\ne-3.$
Work Step by Step
The product of a binomial sum and a binomial difference is: $(A+B)(A-B)=A^2-B^2$.
Hence:
$\frac{x^2-49}{x^2-4x-21}\frac{x+3}{x}=\frac{x^2-7^2}{x^2+3x-7x-21}\frac{x+3}{x}=\frac{(x+7)(x-7)}{x(x+3)-7(x+3)}=\frac{(x+7)(x-7)}{(x+3)(x-7)}\frac{x+3}{x}=\frac{x+7}{x}$ and $x\ne7, x\ne-3.$
