Đáp án:
$\\$
`(x-1)/2009 + (x-2)/2008 = x/1005`
Trừ cả 2 vế với `2` ta được :
`-> (x-1)/2009 + (x-2)/2008 - 2 = x/1005 - 2`
`-> (x-1)/2009 + (x-2)/2008 - 1-1 =x/1005 - 2010/1005`
`-> ( (x-1)/2009 - 1) + ( (x-2)/2008 - 1 ) = (x-2010)/1005`
`-> ( (x-1)/2009 - 2009/2009) + ( (x-2)/2008 - 2008/2008) = (x-2010)/1005`
`-> (x-1-2009)/2009+ (x-2-2008)/2008 = (x-2010)/1005`
`-> (x-2010)/2009 + (x-2010)/2008 = (x-2010)/1005`
`-> (x-2010)/2009 + (x-2010)/2008 - (x-2010)/1005`
`-> (x-2010) (1/2009 + 1/2008 - 1/1005)=0`
`-> x-2010=0` (Do `1/2009 + 1/2008 - 1/1005 \ne 0`)
`->x=0+2010`
`->x=2010`
Vậy `x=2010`
