Giải:
Ta có:
\(ab-ac+bc-c^2=-1\)
\(\Leftrightarrow a\left(b-c\right)+c\left(b-c\right)=-1\)
\(\Leftrightarrow\left(b-c\right)\left(a+c\right)=-1\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}b-c=1\\a+c=-1\end{matrix}\right.\\\left\{{}\begin{matrix}b-c=-1\\a+c=1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(b-c\right)+\left(a+c\right)=1+\left(-1\right)\\\left(b-c\right)+\left(a+c\right)=\left(-1\right)+1\end{matrix}\right.\)
\(\Leftrightarrow b+a=0\)
\(\Leftrightarrow a;b\) là hai số nguyên tố cùng nhau
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{a}{b}=\dfrac{-a}{a}=-1\\\dfrac{a}{b}=\dfrac{a}{-a}=-1\end{matrix}\right.\)
Vậy \(\dfrac{a}{b}=-1\)
