Đáp án:

$\begin{array}{l}
\dfrac{1}{2} - \left( {\dfrac{1}{3} + \dfrac{1}{4}} \right) < x < \dfrac{1}{{48}} - \left( {\dfrac{1}{{16}} - \dfrac{1}{6}} \right)\\
 \Leftrightarrow \dfrac{1}{2} - \dfrac{1}{3} - \dfrac{1}{4} < x < \dfrac{1}{{48}} - \dfrac{1}{{16}} + \dfrac{1}{6}\\
 \Leftrightarrow \dfrac{{6 - 4 - 3}}{{12}} < x < \dfrac{{1 - 3 + 8}}{{48}}\\
 \Leftrightarrow \dfrac{{ - 1}}{{12}} < x < \dfrac{6}{{48}}\\
 \Leftrightarrow  - \dfrac{1}{{12}} < x < \dfrac{1}{8}\\
Vậy\, - \dfrac{1}{{12}} < x < \dfrac{1}{8}
\end{array}$

Đáp án:

Giải thích các bước giải:

`1/2 - (1/3 + 1/4) < x < 1/48 - (1/16 -1/6)`

`<=> 1/2 - 1/3 - 1/4 < x < 1/48 - 1/16 + 1/6`

`<=> 6/12 - 4/12 - 3/12 < x < 1/48 - 3/48 + 8/48`

`=> -1/2 < x < 1/8` 

Vậy `-1/2 < x < 1/8` (nếu đề bảo `x ∈ Z` thì `x = 0`)