\(\left(1\dfrac{2}{25}+\dfrac{14}{9}+\dfrac{17}{5}\right)-\left(\dfrac{2}{25}+\dfrac{5}{9}+40\%\right)\)
\(=\left(1+\dfrac{2}{25}+\dfrac{14}{9}+\dfrac{17}{5}\right)-\left(\dfrac{2}{25}+\dfrac{5}{9}+\dfrac{40}{100}\right)\)
\(=1+\dfrac{2}{25}+\dfrac{14}{9}+\dfrac{17}{5}-\left(\dfrac{2}{25}+\dfrac{5}{9}+\dfrac{2}{5}\right)\)​
\(=1+\dfrac{2}{25}+\dfrac{14}{9}+\dfrac{17}{5}-\dfrac{2}{25}-\dfrac{5}{9}-\dfrac{2}{5}\)
\(=1+\left(\dfrac{2}{25}+\dfrac{-2}{25}\right)+\left(\dfrac{14}{9}+\dfrac{-5}{9}\right)+\left(\dfrac{17}{5}+\dfrac{-2}{5}\right)\)
\(=1+0+1+3\)
\(=5\)​
 

\(\dfrac{x}{6}-\dfrac{2}{y}=\dfrac{1}{30}\)
\(\dfrac{xy}{6y}-\dfrac{12}{6y}=\dfrac{1}{30}\)
\(\dfrac{xy-12}{6y}=\dfrac{1}{30}\)
\(\Rightarrow\left(xy-12\right).30=6y.1\)
\(\left(xy-12\right).30=6y\)
\(xy-12=6y:30\)
\(\Rightarrow xy-12=\dfrac{6y}{30}\)
\(xy-12=\dfrac{y}{5}\)
\(xy-12=y:5\)
\(xy-12=y.\dfrac{1}{5}\)
\(xy=y.\dfrac{1}{5}+12\)
\(xy-y.\dfrac{1}{5}=12\)
\(y.\left(x-\dfrac{1}{5}\right)=12\)
\(\Rightarrow\) \(y\) và \(\left(x-\dfrac{1}{5}\right)\) là ước của 12
Ước của 12 là:\(1;-1;2;-2;3;-3;4;-4;6;-6;12;-12.\)
Ta có bảng:
 

\(\Rightarrow\) Không có cặp nào thoả mãn.
Vậy có 0 cặp số nguyên (x;y) thoả mãn \(\dfrac{x}{6}-\dfrac{2}{y}=\dfrac{1}{30}\).