Giải:
Ta có: \(\dfrac{6}{5}< x-\dfrac{3}{2}< \dfrac{12}{5}\)
\(\Leftrightarrow\dfrac{12}{10}< \dfrac{10x}{10}-\dfrac{15}{10}< \dfrac{24}{10}\)
\(\Leftrightarrow12< 10x-15< 24\)
\(\Leftrightarrow12< 10x< 39\)
\(\Leftrightarrow10x=\left\{13;14;15;...;37;38\right\}\)
Mà \(x\in Z\)
\(\Leftrightarrow10x=\left\{20;30\right\}\)
\(\Leftrightarrow x=\left\{2;3\right\}\)
Vậy ...