`#khanhly2949`

`(x-1)/2015+(x-3)/2013=(x-5)/2011+(x-7)/2009`

`<=>(x-1)/2015-1+(x-3)/2013-1=(x-5)/2011-1+(x-7)/2009-1`

`<=>(x-2016)/2015+(x-2016)/2013=(x-2016)/2011+(x-2016)/2009`

`<=>(x-2016)/2015+(x-2016)/2013-(x-2016)/2011-(x-2016)/2009=0`

`<=>(x-2016) . (1/2015+1/2013-1/2011-1/2009)=0`

`<=>x-2016=0`

`<=>x=2016`

Vậy `x=2016`