`(x+1)/2008+(x+2)/2007=(x-1)/2010+(x-2)/2011`
`⇔(x+1)/2008+(x+2)/2007-(x-1)/2010-(x-2)/2011=0`
`⇔(x+1)/2008+1+(x+2)/2007+1-((x-1)/2010+1)-((x-2)/2011+1)=0`
`⇔(x+1+2008)/2008+(x+2+2007)/2007-(x-1+2010)/2010-(x-2+2011)/2011=0`
`⇔(x+2009)/2008+(x+2009)/2007-(x+2009)/2010-(x+2009)/2011=0`
`⇔(x+2009)(1/2008+1/2007-1/2010-1/2011)=0`
`⇔x+2009=0`
`⇔x=-2009`
Vì `1/2008+1/2007-1/2010-1/2011 \ne 0`