Bài giải:

$\begin{align}& A=\frac{\left( 1+\frac{2018}{1} \right)\times \left( 1+\frac{2018}{2} \right)\times \left( 1+\frac{2018}{3} \right)\times ...\times \left( 1+\frac{2018}{1000} \right)}{\left( 1+\frac{1000}{1} \right)\times \left( 1+\frac{1000}{2} \right)\times \left( 1+\frac{1000}{3} \right)\times ...\times \left( 1+\frac{1000}{2018} \right)} \\  & =\frac{\frac{2019}{1}\times \frac{2020}{2}\times \frac{2021}{3}\times ...\times \frac{3018}{1000}}{\frac{1001}{1}\times \frac{1002}{2}\times \frac{1003}{3}\times ...\times \frac{3018}{2018}} \\ & =\frac{2019\times 2020\times 2021\times ...\times 3018}{1\times 2\times 3\times ...\times 1000}\times \frac{1\times 2\times 3\times ...\times 2018}{1001\times 1002\times 1003\times ...\times 3018} \\  & =\frac{1\times 2\times 3\times ...\times 2018\times 2019\times 2020\times ...\times 3018}{1\times 2\times 3\times ...\times 2018\times 2019\times 2020\times ...\times 3018}=1 \\ \end{align}$