Question

tính:

\(\left(\dfrac{1}{10}-1\right).\left(\dfrac{1}{11}-1\right).\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)

Answer 1

\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)

\(=\dfrac{-9}{10}.\dfrac{-10}{11}.\dfrac{-11}{12}....\dfrac{-98}{99}.\dfrac{-99}{100}\)

\(=\dfrac{-9.-10.-11.....-98.-99}{10.11.12....99.100}\)

Vì :

Từ \(-9\) đến \(-99\) có :

\(\left(99-9\right):1+1=91\) số hạng.

Nên:

\(\)\(\dfrac{-9.-10.-11.....-98.-99}{10.11.12....99.100}=\dfrac{-\left(9.10.11....98.99\right)}{10.11.12.99.100}\)

\(=\dfrac{-9}{100}\)