Đáp án:
$\\$
`1 + 1/3 + ... + 2/(x (x+1) ) = 1 2019/2021` (Điều kiện : `x\ne 0, x\ne -1`)
`-> 2/2 + 2/6 + ... + 2/(x (x+1) ) = 4040/2021`
`-> 2 × [1/2 + 1/6 + ... + 1/(x (x+1) )] = 4040/2021`
`-> 2 × [1/(1×2) +1/(2×3) + ... + 1/(x (x+1) ) ]=4040/2021`
`-> 2 × [1 - 1/2 + 1/2 - 1/3 + ... + 1/x - 1/(x+1)]=4040/2021`
`-> 2 × [1 + (-1/2 + 1/2) + ... + (-1/x + 1/x) - 1/(x+1)]=4040/2021`
`-> 2 × [1 - 1/(x+1)]=4040/2021`
`-> 1 - 1/(x+1) = 4040/2021 ÷ 2`
`-> 1- 1/(x+1) = 2020/2021`
`-> 1/(x+1) = 1 -2020/2021`
`-> 1/(x+1) = 2021/2021 - 2020/2021`
`-> 1/(x+1) = 1/2021`
`->x+1=2021`
`->x=2021-1`
`->x=2020` (Thỏa mãn)
Vậy `x=2020`
