\(S=\frac{1}{1+a+ab}+\frac{1}{1+b+bc}+\frac{1}{1+c+ac}\)
=\(\frac{c}{c\left(1+a+ab\right)}+\frac{ac}{ac\left(1+b+bc\right)}+\frac{1}{1+c+ac}\)
=\(\frac{c}{c+ac+abc}+\frac{ac}{ac+abc+abc.c}+\frac{1}{1+c+ac}\)
thay abc=1 ta được:
\(\frac{c}{c+ac+1}+\frac{ac}{ac+1+c}+\frac{1}{1+c+ac}\)(cùng mẫu c+ac+1)
=\(\frac{c+ac+1}{c+ac+1}=1\)
vậy S=1