Đáp án:
$\\$
`1/15 + 1/21 + ... + 2/(x (x+1) ) = 11/40` (Điều kiện : `x \ne 0; x\ne -1`)
`-> 2/30 + 2/42 + ... + 2/(x (x+1) ) = 11/40`
`-> 2/(5×6) + 2/(6×7) + ... + 2/(x (x+1) ) = 11/40`
`-> 2 × [1/(5×6) + 1/(6×7) + ... + 1/(x(x+1) )]=11/40`
`-> 2 × [1/5 - 1/6 + 1/6 - 1/7 + ... + 1/x - 1/(x+1) ]=11/40`
`-> 2 × [1/5 + (-1/6 + 1/6) + ... + (-1/x +1/x) - 1/(x+1)]=11/40`
`-> 2 × [1/5 - 1/(x+1)]=11/40`
`-> 1/5 - 1/(x+1) = 11/40 ÷2`
`-> 1/5 - 1/(x+1)=11/80`
`-> 1/(x+1)=1/5 - 11/80`
`->1/(x+1)=1/16`
`->x+1=16`
`->x=16-1`
`->x=15` (tm)
Vậy `x=15`
