Câu trả lời:
a) A=\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{99.100}\)
A= 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)
A= 1-\(\dfrac{1}{100}\)
A= \(\dfrac{99}{100}\)