\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{2a}{2b}=\dfrac{3c}{3d}=\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

\(\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{999}}{\dfrac{1}{1.999}+\dfrac{1}{3.997}+...+\dfrac{1}{997.3}+\dfrac{1}{999.1}}\)

\(=\dfrac{\left(1+\dfrac{1}{999}\right)+\left(\dfrac{1}{3}+\dfrac{1}{997}\right)+...+\left(\dfrac{1}{499}+\dfrac{1}{501}\right)}{2\left(\dfrac{1}{1.999}+\dfrac{1}{3.997}+...+\dfrac{1}{499.501}\right)}\)

\(=\dfrac{\dfrac{1000}{1.999}+\dfrac{1000}{3.997}+...+\dfrac{1000}{499.501}}{2\left(\dfrac{1}{1.999}+\dfrac{1}{3.997}+...+\dfrac{1}{499.501}\right)}\)

\(=\dfrac{1000\left(\dfrac{1}{1.999}+\dfrac{1}{3.997}+...+\dfrac{1}{499.501}\right)}{2\left(\dfrac{1}{1.999}+\dfrac{1}{3.997}+...+\dfrac{1}{499.501}\right)}\)

\(=\dfrac{1000}{2}=500\)