Bài giải:

$\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}$

$\Leftrightarrow \frac{2-x}{2001}-1+2=\frac{1-x}{2002}-\frac{x}{2003}+2$

$\Leftrightarrow \frac{2-x}{2001}+1=\left( \frac{1-x}{2002}+1 \right)-\left( \frac{x}{2003}-1 \right)$

$\Leftrightarrow \frac{2-x+2001}{2001}=\frac{1-x+2002}{2002}-\frac{x-2003}{2003}$

$\Leftrightarrow \frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}$

$\Leftrightarrow \frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0$

$\Leftrightarrow (2003-x)\left( \frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003} \right)=0$

Mà:  $\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne 0$

Nên 2003 – x = 0

$\Rightarrow $ x = 2003

Vậy tập nghiệm của  phương trình là:S = {2003}