$$\begin{gathered} 1. \left(1+\frac{1}{51}\right)\left(1+\frac{1}{52}\right)...\left(1+\frac{1}{1963}\right) \\ =\frac{52}{51}.\frac{53}{52}....\frac{1964}{1963} \\ =\frac{1964}{51} \end{gathered}$$..

Câu `1`

`(1 + 1/51) xx (1+1/52) xx (1+1/53) xx ... xx (1+1/1963)`

`= 52/51 xx 53/52 xx 54/53 xx ... xx 1964/1963`

`= (52xx53xx54xx...xx1964)/(51xx52xx53xx...xx1963)`

`= 1964/51`

Câu `2`

`(1-1/400) xx (1-1/401) xx (1-1/402) xx ... xx (1-1/2022)`

`= 399/400 xx 400/401 xx 401/402 xx ... xx 2021/2022`

`= (399xx400xx401 xx ... xx 2021)/(400 xx 401 xx 402 xx ... xx 2022)`

`= 399/2022`

Câu `3`

`1/(2xx4) + 1/(4xx7) + 1/(7xx10) + ... + 1/(37xx40)`

`= 1/6 + 1/3 xx (3/(4xx7) + 3/(7xx10) + ... + 3/(37xx40))`

`= 1/6 + 1/3 xx (1/4 - 1/7 + 1/7 - 1/10 + ... + 1/37 - 1/40)`

`= 1/6 + 1/3 xx (1/4 - 1/40)`

`= 1/6 + 1/3 xx 9/40`

`= 1/6 + 9/120`

`= 29/120`

$\frac{1964}{51}$

câu 2 bạn có ghi sai đề không ak

`1)(1+1/51)xx(1+1/52)xx...xx(1+1/1963)~`

`=52/51xx53/52xx...xx1964/1963`

`=1964/51`

`2)(1-1/400)xx(1-1/401)xx...xx(1-1/2022)`

`=399/400x400/401xx...xx2021/2022`

`=399/2022`

`=133/674`

`3)1/(1xx4)+1/(4xx7)+...+1/(37xx40)`

`=1/3(3/(1xx4)+...+3/(37xx40))`

`=1/3xx(1-1/4+...+1/37-1/40)`

`=1/3xx(1-1/40)`

`=1/3xx(40/40-1/40)`

`=1/3xx39/40`

`=13/40`