Ta có:
`(x+1)/2009+(x+2)/2008=(x+2007)/3+(x+2006)/4`
`<=>((x+1)/2009+1)+((x+2)/2008+1)=((x+2007)/3+1)+((x+2006)/4+1)`
`<=>(x+2010)/2009+(x+2010)/2008=(x+2010)/3+(x+2010)/4`
`<=>(x+2010)/2009+(x+2010)/2008-(x+2010)/3-(x+2010)/4=0`
`<=>(x+2010)(1/2009+1/2008-1/3-1/4)=0` `(1)`
Lại có:
`1/2009<1/4=>1/2009-1/4<0`
`1/2008<1/3=>1/2008-1/3<0`
`=>1/2009+1/2008-1/3-1/4<0`
Nên `(1)` xảy ra `<=>x+2010=0`
`<=>x=-2010`
Vậy phương trình có nghiệm duy nhất `S={-2010}`
