(1/2-1).(1/3-1).(1/4-1).........(1/2019-1)

\(A=\left({\dfrac{1}{2}-1}\right)\left({\dfrac{1}{3}-1}\right)\left({\dfrac{1}{4}-1}\right)...\left({\dfrac{1}{2019}-1}\right)\)

\(=\left({\dfrac{-1}{2}}\right)\left({\dfrac{-2}{3}}\right)\left({\dfrac{-3}{4}}\right)...\left({\dfrac{-2018}{2019}}\right)\)

\(={(-1)}^{2018}\dfrac{1}{2019}\)

\(=\dfrac{1}{2019}\)

Đáp án:`A=1/2019`.

Giải thích các bước giải:

Đặt `A=(1/2-1).(1/3-1).(1/4-1)......(1/2019-1)`

`A=(1/2-2/2).(1/3-3/3).(1/4-4/4).....(1/2019-2019/2019)`

`A=(-1/2).(-2/3).(-3/4).......(-2018/2019)`

`A=(-1.(-2).(-3)....(-2018))/(2.3.4......2019)`

Từ `-1` đến `-2018` có 2018 số số hạng

Nên `-1.(-2).(-3)....(-2018)=1.2.3.....2018`

`=>A=(1.2.3......2018)/(2.3.4.......2019)=1/2019`.