\(\dfrac{2008a}{ab+2008a+2008}+\dfrac{b}{bc+b+2008}+\dfrac{c}{ca+c+1}=1\)
=>\(\dfrac{2008a}{ab+2008a+2008}+\dfrac{ab}{abc+ab+a2008}+\dfrac{abc}{abca+abc+ab1}=1\)
=>\(\dfrac{2008a}{ab+2008a+2008}+\dfrac{ab}{2008+ab+2008a}+\dfrac{2008}{2008a+2008+ab}=1\)(do abc=2008_
=>\(\dfrac{2008a+2008+ab}{2008a+2008+ab}=1\)