Ta có : \(\frac{ab}{c}+\frac{bc}{a}\in Z\Leftrightarrow\left(\frac{ab}{c}+\frac{bc}{a}\right).c\in\&Z\left(\frac{ab}{c}+\frac{bc}{a}\right).a\in Z\)
\(\Leftrightarrow\hept{\begin{cases}ab+\frac{bc^2}{a}\in Z\\\frac{a^2b}{c}+bc\in Z\end{cases}}a;b;c\in Z\Leftrightarrow\hept{\begin{cases}\frac{bc^2}{a}\in Z\\\frac{a^2b}{c}\in Z\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}bc^2⋮a\\a^2b⋮c\end{cases}\Leftrightarrow a^2b^2c^2⋮ac\Leftrightarrow}b^2⋮ac\Leftrightarrow\hept{\begin{cases}b^2⋮a\\b^2⋮c\end{cases}\Leftrightarrow\hept{\begin{cases}b⋮a\\b⋮c\end{cases}}}\)( nếu a;b;c nguyên tố cùng nhau thì \(b^2\)không \(⋮a;c\))
\(\Rightarrow b=a.k=c.h\left(k;h\in Z\right)\Leftrightarrow\frac{ab}{c}=\frac{a.c.h}{c}=a.h\in Z;\frac{bc}{a}=\frac{a.k.c}{a}=k.c\in Z\)
Vậy \(\frac{ab}{c}+\frac{bc}{a}\in Z\Rightarrow\frac{ab}{c}\in Z;\frac{bc}{a}\in Z\left(đpcm\right).\)
