Đáp án:
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`1 + 1/3 + 1/6 + 1/10 + ... +2/(x (x + 1) ) = 1 1989/1991`
`↔ 2/2 + 2/6 + 2/12 + 2/20 + ... + 2/(x (x + 1) ) = 3980/1991`
`↔ 2 × [1/2 + 1/6 + ... + 1/(x (x+1) )] = 3980/1991`
`↔ 2 × [1/(1×2) + 1/(2×3) + ... + 1/(x (x+1) )] = 3980/1991`
`↔ 2 × [1 - 1/2 + 1/2 - 1/3 + ... + 1/x - 1/(x+1)] = 3980/1991`
`↔ 2 × [1 + (- 1/2 + 1/2 - 1/3 + ... + 1/x) - 1/(x+1)] = 3980/1991`
`↔ 2 × [1 - 1/(x+1)] = 3980/1991`
`↔ 1 - 1/(x+1) = 3980/1991 ÷ 2`
`↔ 1 - 1/(x+1) = 3980/1991 × 1/2`
`↔ 1 - 1/(x+1) = 1990/1991`
`↔ 1/(x+1) =1-1990/1991`
`↔1/(x+1)=1991/1991 - 1990/1991`
`↔1/(x+1) = 1/1991`
`↔ x+1=1991`
`↔x=1991-1`
`↔x=1990`
Vậy `x=1990`
