Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
Thêm bớt \(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\) vào A ta được:
\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)
\(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Vậy \(\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}}=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}=1\)
