tính giá trị A =(1+1/3).(1+1/8).(1+1/15)…….(1+1/2021.2023)

2 bình luận về “tính giá trị A =(1+1/3).(1+1/8).(1+1/15)…….(1+1/2021.2023)”

1. (1+$\frac{1}{3}$ ).(1+$\frac{1}{8}$).(1+$\frac{1}{15}$)…..(1+$\frac{1}{2021.2023}$)

   =$\frac{4}{3}$.$\frac{9}{8}$.$\frac{16}{15}$…..$\frac{4088483}{4088484}$

   =$\frac{2.2}{1.3}$.$\frac{3.3}{2.4}$…….$\frac{2022.2022}{2021.2023}$

   =$\frac{2.3.4.5….2022}{1.2…2021}$.$\frac{2.3.4.5….2022}{3.4….2023}$

   =2022.$\frac{2}{2023}$ 

   =$\frac{4044}{2023}$ 

   Trả lời
2. A=(1+1/3)(1+1/8)(1+1/15)……..(1+1/(2021.2023))

   = 4/(1.3) . 9/(2.4) . 16/(3.5) ……….. 4088484/(2021.2023)

   = ((2.3.4……..2022)(2.3.4……..2022))/((1.2.3……2021).(3.4.5…….2023))

   = 2022 . 2/2023

   = 4044/2023

   Trả lời