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`
