$\begin{array}{l}\text{- Ta có : $\dfrac ab=\dfrac{-13}{19}$}\\\to\begin{cases} a=-13k\\b=19k\end{cases}(k\in\mathbb{Z};k\neq0)\\\text{- Ta lại có : $a+b=18$}\\\to -13k+19k=18\\\to k.(-13+19)=18\\\to k.6=18\\\to k=3\\\text{- Thay $k=3$ ta có :}\\\begin{cases} a=-13.3=-39\\b=19.3=57\end{cases}\\\text{- Vậy $\dfrac ab=\dfrac{-39}{57}$} \end{array}$
