Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.10}\)\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)\(=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)Thêm bớt vào tổng trên đại lượng \(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\) ta được:\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}+\dfrac{1}{51}+...+\dfrac{1}{100}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)\(\Rightarrow B=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}=1\)Câu trả lời: Đặt \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.10}\)\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)\(=1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)Thêm bớt vào tổng trên đại lượng \(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\) ta được:\(A=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}-2.\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}+\dfrac{1}{51}+...+\dfrac{1}{100}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)\(\Rightarrow B=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}=1\)