Ta đặt \(\dfrac{a-b\sqrt{3}}{b-c\sqrt{3}}=\dfrac{m}{n}\left(m,n\in Z;m,n\ne0\right)\)
\(\Leftrightarrow an-bn\sqrt{3}=bm-cm\sqrt{3}\)
\(\Leftrightarrow an-bm=\sqrt{3}\left(bn-cm\right)\)(1)
Ta có \(an-bm,bn-cm\) là 2 số nguyên nên để (1) đúng thì
\(\Rightarrow\left\{{}\begin{matrix}an=bm\\bn=cm\end{matrix}\right.\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{b}{c}\)
\(\Rightarrow b^2=ac\)