a: \(A=\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{998}+\dfrac{1}{1000}}{\dfrac{1}{2\cdot1000}+\dfrac{1}{4\cdot998}+...+\dfrac{1}{998\cdot4}+\dfrac{1}{1000\cdot2}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{998}+\dfrac{1}{1000}}{\dfrac{2}{2\cdot1000}+\dfrac{2}{4\cdot998}+...+\dfrac{2}{500\cdot502}}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{998}+\dfrac{1}{1000}}{\dfrac{2}{1002}\left(\dfrac{1002}{2\cdot1000}+\dfrac{1002}{4\cdot998}+...+\dfrac{1002}{500\cdot502}\right)}\)
\(=\dfrac{\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{998}+\dfrac{1}{1000}}{\dfrac{1}{501}\left(\dfrac{1}{2}+\dfrac{1}{1000}+\dfrac{1}{4}+\dfrac{1}{998}+...+\dfrac{1}{500}+\dfrac{1}{502}\right)}=1:\dfrac{1}{501}=501\)
b: \(B=\dfrac{1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{999}}{\dfrac{1}{1\cdot999}+\dfrac{1}{3\cdot997}+...+\dfrac{1}{999\cdot1}}\)
\(=\dfrac{1+\dfrac{1}{3}+...+\dfrac{1}{999}}{\dfrac{2}{1\cdot999}+\dfrac{2}{3\cdot997}+...+\dfrac{2}{499\cdot501}}\)
\(=\dfrac{1+\dfrac{1}{3}+...+\dfrac{1}{999}}{\dfrac{1}{500}\cdot\left(\dfrac{1000}{1\cdot999}+\dfrac{1000}{3\cdot997}+...+\dfrac{1000}{499\cdot501}\right)}\)
\(=\dfrac{1+\dfrac{1}{3}+...+\dfrac{1}{999}}{\dfrac{1}{500}\left(1+\dfrac{1}{999}+\dfrac{1}{3}+\dfrac{1}{997}+...+\dfrac{1}{499}+\dfrac{1}{501}\right)}=1:\dfrac{1}{500}=500\)
c: \(C=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}}\)
\(=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{1+\dfrac{1}{3}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{1+\dfrac{1}{2}+...+\dfrac{1}{99}+\dfrac{1}{100}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{1+\dfrac{1}{2}+...+\dfrac{1}{99}+\dfrac{1}{100}-1-\dfrac{1}{2}-...-\dfrac{1}{50}}\)
\(=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}=1\)
d: \(D=\dfrac{\dfrac{101}{1}+\dfrac{100}{2}+\dfrac{99}{3}+...+\dfrac{1}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{102}}\)
\(=\dfrac{\left(\dfrac{100}{2}+1\right)+\left(\dfrac{99}{3}+1\right)+...+\left(\dfrac{1}{101}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{102}}\)
\(=\dfrac{\dfrac{102}{2}+\dfrac{102}{3}+...+\dfrac{102}{101}+\dfrac{102}{101}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{102}}=102\)
