HÃy tìm số nguyên x biết ​\(\begin{aligned} &\frac{2}{3} \cdot\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right) \leq \frac{x}{18} \leq \frac{7}{3} \cdot\left(\frac{1}{2}-\frac{1}{6}...

* Hướng dẫn giải

 \(\begin{aligned} &\begin{aligned} &\text { Ta có: } \frac{2}{3} \cdot\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)=\frac{2}{3} \cdot\left(\frac{6}{12}+\frac{9}{12}-\frac{4}{12}\right)=\frac{2}{3} \cdot \frac{11}{12}=\frac{11}{18} \\ &\frac{7}{3} \cdot\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{7}{3} \cdot\left(\frac{6}{12}-\frac{2}{12}\right)=\frac{7}{3} \cdot \frac{4}{12}=\frac{7}{3} \cdot \frac{1}{3}=\frac{7}{9}=\frac{14}{18} . \end{aligned}\\ &\text {Khi đó } \frac{11}{18} \leq \frac{x}{18} \leq \frac{14}{18} \text { nên } x \in \mathbb{Z} \text { . Suy ra } x \in\{11 ; 12 ; 13 ; 14\} \text { . } \end{aligned}\)