Hãy tìm số nguyên x biết ​\(\left(\frac{5}{12}+\frac{3}{8}-\frac{2}{3}\right) \cdot \frac{-2}{3}\le \frac{x}{24} \le \frac{1}{4} \cdot \frac{5}{2}+\frac{1}{6} \cdot \frac{5}{2}-\fr...

* Hướng dẫn giải

 \(\begin{aligned} &\begin{aligned} &\text {Ta có: }\left(\frac{5}{12}+\frac{3}{8}-\frac{2}{3}\right) \cdot \frac{-2}{3}=\left(\frac{10}{24}+\frac{9}{24}-\frac{16}{24}\right) \cdot \frac{-2}{3}=\frac{3}{24} \cdot \frac{-2}{3}=\frac{-2}{24} \\ &\frac{1}{4} \cdot \frac{5}{2}+\frac{1}{6} \cdot \frac{5}{2}-\frac{5}{2} \cdot \frac{1}{3}=\frac{5}{2} \cdot\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{3}\right)=\frac{5}{2} \cdot\left(\frac{3}{12}+\frac{2}{12}-\frac{4}{12}\right)=\frac{5}{2} \cdot \frac{1}{12}=\frac{5}{24} . \end{aligned}\\ &\text {Khi đó: }\frac{-2}{24}<\frac{x}{24}<\frac{5}{24} \text { và } x \in \mathbb{Z} \text { , suy ra } x \in\{-1 ; 0 ; 1 ; 2 ; 3 ; 4\} \text { . } \end{aligned}\)