Câu trả lời:
\(\dfrac{1}{2.15}\)+\(\dfrac{1}{15.3}\)+\(\dfrac{1}{3.21}\)+...........+\(\dfrac{6}{87.90}\)
= 6.(\(\dfrac{1}{12.15}\)+\(\dfrac{1}{15.18}\)+\(\dfrac{1}{18.21}\)+.............+\(\dfrac{1}{87.90}\))
= \(\dfrac{6}{3}\).(\(\dfrac{1}{12}\)−\(\dfrac{1}{15}\)+\(\dfrac{1}{15}\)−\(\dfrac{1}{18}\)+\(\dfrac{1}{18}\)−\(\dfrac{1}{21}\)+..........+\(\dfrac{1}{87}\)−\(\dfrac{1}{90}\))
= \(\dfrac{6}{3}\).(\(\dfrac{1}{12}\)−\(\dfrac{1}{90}\))
= \(\dfrac{6}{3}\).\(\dfrac{13}{180}\)
= \(\dfrac{13}{90}\)