tồn tại A => a,b,c khác 0
=> a+b+c=0
\(A=\left(\dfrac{a}{b}+1.\right)\left(\dfrac{b}{c}+1.\right)\left(.\dfrac{c}{a}+1\right)=\left(\dfrac{a+b}{b}\right).\left(\dfrac{b+c}{c}\right).\left(\dfrac{c+a}{a}\right)=\left(\dfrac{-c}{b}\right).\left(\dfrac{-a}{c}\right).\left(\dfrac{-b}{a}\right)=-1\)\(\)