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

$\dfrac{a}{m} + \dfrac{b}{m} = \dfrac{{a - b}}{m}$ 

$\dfrac{a}{m} + \dfrac{b}{m} = \dfrac{{a.b}}{m}$  

$\dfrac{a}{m} + \dfrac{b}{m} = \dfrac{{a + b}}{m}$        

$\dfrac{a}{m} + \dfrac{b}{m} = \dfrac{{a + b}}{{m + m}}$

\(\dfrac{{ - 13}}{7}\)

\(\dfrac{{13}}{{ - 7}}\)           

\( - \dfrac{{13}}{7}\)         

Tất cả các đáp án trên đều đúng

$\dfrac{1}{3}$                     

\(\dfrac{4}{3}\)  

\(\dfrac{3}{4}\)                 

\(1\)

$\dfrac{1}{{10}}$ 

$\dfrac{4}{5}$           

\(\dfrac{2}{5}\)          

\(\dfrac{{ - 1}}{{10}}\)

$\dfrac{{ - 5}}{2}$ 

\( - \dfrac{{29}}{5}\)  

\(\dfrac{{ - 40}}{9}\)

\(\dfrac{{40}}{9}\)

\(\dfrac{73}{84}\)

\(\dfrac{-13}{84}\)

\(\dfrac{83}{84}\)

\(\dfrac{143}{84}\)

$\dfrac{{ - 4}}{{11}} + \dfrac{7}{{ - 11}} > 1$ 

$\dfrac{{ - 4}}{{11}} + \dfrac{7}{{ - 11}} < 0$  

$\dfrac{8}{{11}} + \dfrac{7}{{ - 11}} > 1$

$\dfrac{{ - 4}}{{11}} + \dfrac{{ - 7}}{{11}} >  - 1$