Bài giải:

$ab-ac+bc-{{c}^{2}}=-1$

$a\left( b-c \right)+c\left( b-c \right)=-1$

$\left( a+c \right)\left( b-c \right)=-1$

TH1.

$\left\{ \begin{align}& a+c=1 \\ & b-c=-1 \\ \end{align} \right.$

$\left\{ \begin{align}& a=1-c \\  & b=c-1 \\ \end{align} \right.$

Suy ra $\frac{a}{b}=-1$

TH2.

$\left\{ \begin{align}& a+c=-1 \\ & b-c=1 \\ \end{align} \right.$

$\left\{ \begin{align}& a=-1-c \\ & b=1+c \\ \end{align} \right.$

Suy ra  $\frac{a}{b}=-1$

Vậy $\frac{a}{b}=-1$.