Giá trị \(x \in \mathbb{Z}\) thoả mãn \(\frac{1}{2} - \left( {\frac{1}{3} + \frac{3}{4}} \right) \le x \le \frac{1} - \left( {\frac{1}{8} - \frac{1}{3}} \right)\) là:

Đáp án đúng là: B.

\(\frac{1}{2} - \left( {\frac{1}{3} + \frac{3}{4}} \right) \le x \le \frac{1} - \left( {\frac{1}{8} - \frac{1}{3}} \right)\)

\(\frac{1}{2} - \left( {\frac{4} + \frac{9}} \right) \le x \le \frac{1} - \left( {\frac{3} - \frac{8}} \right)\)

\(\frac{1}{2} - \frac \le x \le \frac{1} - \frac{{ - 5}}\)

\(\frac{6} - \frac \le x \le \frac{1} + \frac{5}\)

\(\frac{{ - 7}} \le x \le \frac{6}\)

\(\frac{{ - 7}} \le x \le \frac{1}{4}\)

Ta có \(\frac{{ - 12}} < \frac{{ - 7}} \le x \le \frac{1}{4} < \frac{4}{4}\)

Hay \( - 1 < \frac{{ - 7}} \le x \le \frac{1}{4} < 1\)

Do đó giá trị x nguyên thoả mãn là x = 0.