Giải phương trình

(x+1)/2009+(x+3)/2007=(x+5)/2005+(x+7)/1993

`Parkchaeyoung`

$\frac{x+1}{2009}$`+` $\frac{x+3}{2007}$`=` $\frac{x+5}{2005}$`+` $\frac{x+7}{2003}$ 

`<=>` $\frac{x+1}{2009}$`+1+` $\frac{x+3}{2007}$`+1=` $\frac{x+5}{2005}$`+1+` $\frac{x+7}{2003}$`+1`

`<=>` $\frac{x+1+2009}{2009}$`+` $\frac{x+3+2007}{2007}$`=` $\frac{x+5+2005}{2005}$`+` $\frac{x+7+2003}{2003}$

`<=>` $\frac{x+2010}{2009}$`+` $\frac{x+2010}{2007}$`-` $\frac{x+2010}{2005}$`-` $\frac{x+2010}{2003}$`=0`

`<=> (x+2010)(1/2009+1/2007-1/2005-1/2003)=0`

`<=> x+2010=0`

`<=> x=-2010`

Vậy PT có tập nghiệm `S={-2010}`