Tìm \(x \in Z\) biết \(\dfrac{5}{6} + \dfrac{{ - 7}}{8} \le \dfrac{x}{{24}} \le \dfrac{{ - 5}}{{12}} + \dfrac{5}{8}\).

$\dfrac{{ - 2}}{3};\dfrac{3}{2}$                 

$\dfrac{{ - 12}}{{13}};\dfrac{{13}}{{ - 12}}$  

\(\dfrac{1}{2}; - \dfrac{1}{2}\)

\(\dfrac{3}{4};\dfrac{{ - 4}}{3}\)

$\dfrac{{ - 53}}{{35}}$ 

\(\dfrac{{51}}{{35}}\)  

\(\dfrac{{ - 3}}{{35}}\)                  

\(\dfrac{3}{{35}}\)

$\dfrac{3}{2} + \dfrac{2}{3} > 1$ 

$\dfrac{3}{2} + \dfrac{2}{3} = \dfrac{{13}}{6}$     

$\dfrac{3}{4} + \left( {\dfrac{{ - 4}}{{17}}} \right) = \dfrac{{35}}{{68}}$

$\dfrac{4}{{12}} + \dfrac{{21}}{{36}} = 1$

$\dfrac{9}{{14}}$ 

\(\dfrac{1}{{14}}\)  

\(\dfrac{{11}}{{14}}\)                 

\(\dfrac{1}{2}\)

\(\dfrac{{12}}{{33}}\) 

\(\dfrac{{177}}{{260}}\)  

\(\dfrac{{187}}{{260}}\)

\(\dfrac{{177}}{{26}}\)

$\dfrac{4}{{13}} - \dfrac{1}{2} = \dfrac{5}{{26}}$                       

\(\dfrac{1}{2} - \dfrac{1}{3} = \dfrac{5}{6}\)  

\(\dfrac{{17}}{{20}} - \dfrac{1}{5} = \dfrac{{13}}{{20}}\)

\(\dfrac{5}{{15}} - \dfrac{1}{3} = \dfrac{1}{5}\)

$M = \dfrac{2}{7};N = \dfrac{1}{{41}}$     

$M = 0;N = \dfrac{1}{{41}}$

\(M = \dfrac{{ - 16}}{7};N = \dfrac{{83}}{{41}}\)

$M =  - \dfrac{2}{7};N = \dfrac{1}{{41}}$