Theo đề bài ta có :
\(\dfrac{1}{3}< \dfrac{x}{y}< \dfrac{2}{5}\)
- Với \(\dfrac{x}{y}< \dfrac{2}{5}\)
\(\Rightarrow\dfrac{x}{2}< \dfrac{y}{5}< \dfrac{55x+53y}{55.2+53.5}=\dfrac{1981}{375}\sim5,3\)
\(\Rightarrow\left\{{}\begin{matrix}x< 2.5,3=10,6\\y< 5.5,3=26,5\end{matrix}\right.\) \(\left(1\right)\)
- Với \(\dfrac{x}{y}>\dfrac{1}{3}\)
\(\Rightarrow\dfrac{x}{1}>\dfrac{y}{3}>\dfrac{55x+53y}{55.1+53.3}=\dfrac{1981}{214}\sim9,3\)
\(\Rightarrow\left\{{}\begin{matrix}x>9,3\\y>3.9,3=27,9\end{matrix}\right.\) \(\left(2\right)\)
\(\left(1\right);\left(2\right)\Rightarrow\left\{{}\begin{matrix}9,3< x< 10,6\\26,6< y< 27,9\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=10\\y=27\end{matrix}\right.\)
Kiểm tra \(55x+53y=55.10+53.27=1981\left(đúng\right)\)
Vậy \(\left(x;y\right)=\left(10;27\right)\) thỏa mãn đề bài
