`1/{3.5} + 1/{5.7} + .... + 1/{(2x+1).(2x+3)} = {15}/{93}`
`⇔ 1/2 . (2/{3.5} + 2/{5.7} + .... + 2/{(2x+1).(2x+3)}) = {15}/{93}`
`⇔ 1/2 . (1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/{2x+1} - 1/{2x+3} ) = {15}/{93}`
`⇔ 1/2 . (1/3 - 1/{2x+3}) = {15}/{93}`
`⇔ 1/3 - 1/{2x+3} = {10}/{31}`
`⇔ 1/{2x+3} = 1/{93}`
`⇔ 2x = 90`
`⇔ x = 45`
Vậy $x=45$