Tính phép Tính sau (1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]

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1. Đặt:

   `A = 1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + … + 1/(99.100)`

   `A = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 + 1/4 – 1/5 + … + 1/99 – 1/100`

   `A = 1 – 1/100`

   `A = 99/100`

   Đặt:

   `B = (1 – 1/2) . (1 – 1/3) . (1 – 1/4) . …  . (1 – 1/2022)`

   `B = 1/2 . 2/3 . 3/4 . …  . 2021/2022`

   `B = (1 . 2 . 3 . …  . 2021)/(2 . 3 . 4 . …  . 2022)`

   `B = 1/2022`

   Như vậy:

   `(1/(1.2) + 1/(2.3) + 1/(3.4) + 1/(4.5) + … + 1/(99.100)) . [(1 – 1/2) . (1 – 1/3) . (1 – 1/4) . …  . (1 – 1/2022)]

   `= 99/100 . 1/2022`

   `= 99/202200`

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2. Cách 1:Chi tiết,dễ hiểu

   `(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]`

   Đặt :
   `A=1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100`
   `B=(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)`
   `=>(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]=A.B`
   Theo bài ra ta có:
   `A=1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100`
   Áp dụng công thức `1/[n.(n+1)]=1/n-1/[n+1]`
   `=>A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/99-1/100`
   `A=1-1/100`
   `A=100/100-1/100`
   `A=99/100`
   Vậy `A=99/100(1)`
   Lại có:
   `B=(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)`
   `B=(2/2-1/2)(3/3-1/3)(4/4-1/4)…(2022/2022-1/2022)`
   `B=1/2. 2/3. 3/4…. 2021/2022`
   `B=1/2022`
   Vậy `B=1/2022(2)`
   Từ `(1),(2)`
   `=>A.B=99/100. 1/2022`
   `=>A.B=99/202200`
   `=>(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]=99/202200`

   Cách 2:Ngắn gọn,hơi khó hiểu với một số bạn.

   `(1/1.2+1/2.3+1/3.4+1/4.5+….+1/99.100).[(1-1/2).(1-1/3)(1-1/4)…..(1-1/2022)]`
   `=(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+…+1/99-1/100).[(2/2-1/2)(3/3-1/3)(4/4-1/4)…(2022/2022-1/2022)]`
   `=(1-1/100).[1/2. 2/3. 3/4…. 2021/2022]`
   `=(100/100-1/100). 1/2022`
   `=99/100. 1/2022`
   `=99/202200`

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