Thay abc = 1 ta có:
\(\dfrac{1}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{abc+bc+b}\)
\(=\dfrac{abc}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{1}{1+bc+b}\)
\(=\dfrac{abc}{a\left(b+1+bc\right)}+\dfrac{b}{bc+b+1}+\dfrac{1}{bc+b+1}\)
\(=\dfrac{bc}{bc+b+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{bc+b+1}\)
\(=\dfrac{bc+b+1}{bc+b+1}=1\)
Vậy \(\dfrac{1}{ab+a+1}+\dfrac{b}{bc+b+1}+\dfrac{1}{abc+bc+1}=1\)