Đáp án:
$\\$
`a,`
`1/3 + 1/15 + ... + 1/(x (x+2) ) = 1012/2025` (Điều kiện : `x\ne 0, x\ne -2`)
`-> 1/(1.3) + 1/(3.5) + ... + 1/(x (x+2) ) = 1012/2025`
`-> 1/2 (1 - 1/3 + 1/3 - 1/5 + ... + 1/x - 1/(x+2) ) = 1012/2025`
`-> 1/2 (1 - 1/(x+2) ) = 1012/2025`
`-> 1 - 1/(x+2) = 1012/2025 : 1/2`
`-> 1 - 1/(x+2)=2024/2025`
`-> 1/(x+2) = 1 - 2024/2025`
`->1/(x+2)=1/2025`
`->x+2=2025`
`->x=2025 - 2`
`->x=2023` (Thỏa mãn)
Vậy `x=2023`
`b,`
`1/3 + 1/6 + ... + 2/(x (x+1) ) = 2021/2023` (Điều kiện : `x \ne 0, x \ne -1`)
`-> 2/6 + 2/12 + ... + 2/(x (x+1) ) = 2021/2023`
`-> 2 (1/6 + 1/12 + ... + 1/(x(x+1) ) )= 2021/2023`
`-> 2 (1/(2.3) +1/(3.4) + ... + 1/(x (x+1) ) ) = 2021/2023`
`-> 2 (1/2 - 1/3 + 1/3 - 1/4 + ... + 1/x - 1/(x+1) ) = 2021/2023`
`-> 2 (1/2 - 1/(x+1) ) = 2021/2023`
`-> 1/2 - 1/(x+1) = 2021/2023 : 2`
`-> 1/2 - 1/(x+1)=2021/4046`
`-> 1/(x+1) = 1/2 - 2021/4046`
`-> 1/(x+1)=1/2023`
`->x+1=2023`
`->x=2023-1`
`->x=2022` (Thỏa mãn)
Vậy `x=2022`
