\(X=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{c+b}{c}\cdot\frac{c+a}{a}\)
Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\c+b=-a\end{cases}}\)
\(\Rightarrow X=\frac{\left(-a\right)\cdot\left(-b\right)\cdot\left(-c\right)}{abc}=-1\)
nên ta đc X là 1 số nguyên