Bài giải:

$\frac{x+1}{2009}+\frac{x+3}{2007}=\frac{x+5}{2005}+\frac{x+7}{1993}$

$\Leftrightarrow \left( \frac{x+1}{2009}+1 \right)+\left( \frac{x+3}{2007}+1 \right)=\left( \frac{x+5}{2005}+1 \right)+\left( \frac{x+7}{1993}+1 \right)$

$\Leftrightarrow \frac{x+2010}{2009}+\frac{x+2010}{2007}-\frac{x+2010}{2005}-\frac{x+2010}{1993}=0$

$\Leftrightarrow (x+2010)\left( \frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{1993} \right)=0$

Vì $\frac{1}{2009}+\frac{1}{2007}-\frac{1}{2005}-\frac{1}{1993}\ne 0$ nên x + 2010 = 0

$\Leftrightarrow $ x = -2010

Vậy nghiệm của phương trình là;  x = -2010