ta có a+b+c=0
=> a+b=-c ; a+c=-b ; b+c=-a
A=\(\left(1+\dfrac{a}{b}\right).\left(1+\dfrac{b}{c}\right).\left(1+\dfrac{c}{a}\right)=\dfrac{b+a}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\) mà b+a=-c ; b+c=-a;a+c=-b
=> A=\(\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=\dfrac{-bca}{bca}=-1\)
Vậy A=-1