ta có :
\(3\frac{1}{3}:2\frac{1}{2}-1\)\(=\frac{10}{3}:\frac{5}{2}-1=\frac{4}{3}-1=\frac{1}{3}\)
\(7\frac{2}{3}.\frac{3}{7}+\frac{5}{2}=\frac{23}{3}.\frac{3}{7}+\frac{5}{2}=\frac{23}{7}+\frac{5}{2}=\frac{81}{14}\)
\(\Rightarrow\frac{1}{3}< x< \frac{81}{14}\)\(\Rightarrow\frac{14}{42}< x< \frac{243}{42}\)
\(\Rightarrow\frac{1}{3}< x< 5\frac{11}{14}\)
mà x\(\in N\Rightarrow x\in\left\{1;2;3;4;5\right\}\)
Vậy x\(\in\left\{1;2;3;4;5\right\}\)