Chứng tỏ rằng:

a. 1-1/2+1/3-1/4+...+1/2021-1/2022=1/1012+1/1013+1/1014+...+1/2022

b.1×3×5×7×..×99= 51/2×52/2×53/2×...× 100/2

`a) 1-1/2 + 1/3 - 1/4 + ... + 1/2021 - 1/2022 = 1/1012 + 1/1013 + 1/1014 + ... + 1/2022`

`1-1/2 + 1/3 - 1/4 + ... + 1/2021 - 1/2022`

`= (1+1/3 + ... + 1/2021) - (1/2 + 1/4 + ... + 1/2022)`

`= (1 + 1/2 + 1/3 +1/4 + ... + 1/2021 + 1/2022) - 2xx(1/2 + 1/4 + ... + 1/2022)`

`= (1+1/2 + 1/3 + 1/4 + ... + 1/2021 + 1/2022) - (1 + 1/2 + 1/3 + ... + 1/1011)`

`= 1/1012 + 1/1013 + 1/1014 + ... + 1/2022`

Vậy

`1-1/2 + 1/3 - 1/4 + ... + 1/2021 - 1/2022 = 1/1012 + 1/1013 + 1/1014 + ... + 1/2022`

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`b)1xx3 xx 5 xx 7 xx ... xx 99 = 51/2 xx 52/2 xx 53/2 xx ... xx 100/2`

`51/2 xx 52/2 xx 53/2 xx ... xx 100/2`

`= (51xx52xx53 xx ... xx 100)/(2^(50))`

`= ((1xx2xx3xx...xx50)xx(51xx52xx53 xx ... xx 100))/((1xx2xx3xx...xx50)xx(2^(50)))`

`= ((1xx2xx3xx...xx100))/((1xx2)xx(2xx2)xx(3xx2)xx...xx(50xx2))`

`= (1xx2xx3xx...xx100)/(2xx4xx6xx...xx100)`

`= 1xx3xx5xx7xx...xx99`

Vậy `1xx3 xx 5 xx 7 xx ... xx 99 = 51/2 xx 52/2 xx 53/2 xx ... xx 100/2`

$@Thwantobethebest$