- Ta có : \(\frac{x-1}{2004}+\frac{x-2}{2003}=\frac{x-3}{2002}+\frac{x-4}{2001}\)
=> \(\frac{x-1}{2004}-1+\frac{x-2}{2003}-1=\frac{x-3}{2002}-1+\frac{x-4}{2001}-1\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}=\frac{x-2005}{2002}+\frac{x-2005}{2001}\)
=> \(\frac{x-2005}{2004}+\frac{x-2005}{2003}-\frac{x-2005}{2002}-\frac{x-2005}{2001}=0\)
=> \(\left(x-2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x-2005=0\)
=> \(x=2005\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2005\right\}\)