Giải ptrinh: `+)“{x+1}/2009+{x+2}/2008={x+2007}/3+{x+2006}/4`

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1. 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}`

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2. `(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]=0`

   `⇒(x+2010)/2009+(x+2010)/2008-(x+2010)/3-(x+2010)/4=0`

   `⇒(x+2010)(1/2009+1/2008-1/3-1/4)=0`

   `⇒x+2010=0⇒x=-2010`

    

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