Tìm \(x \in \mathbb{Z},\) biết:

h) \[\frac{{x - 2}}{{27}} = \frac{3}{{x - 2}}.\]

j) \(5 \cdot \frac{1}{6} + \frac{1}{6} \le x \le \frac{1}{4}:\frac{1}{{13}} + \frac{7}{4}\)

\[\left( {5 + 1} \right) \cdot \frac{1}{6} \le x \le \frac{1}{4} \cdot \frac{{13}}{1} + \frac{7}{4}\]

\[\frac{1}{6} \cdot 6 \le x \le \frac{{13}}{4} + \frac{7}{4}\]

\[1 \le x \le \frac{{20}}{4}\]

\[1 \le x \le 5\]

Vì \[x \in \mathbb{Z}\] nên \[x \in \left\{ {1;\,\,2;\,\,3;\,\,4;\,\,5} \right\}\].

n) \(\left( {2x + \frac{{ - 3}}{4}} \right):\frac{2}{5} = \frac{{ - 10}}{{12}}\)

\[2x + \frac{{ - 3}}{4} = \frac{{ - 10}}{{12}} \cdot \frac{2}{5}\]

\[2x + \frac{{ - 3}}{4} = \frac{{ - 1}}{3}\]

\[2x = \frac{{ - 1}}{3} - \frac{{ - 3}}{4}\]

\[2x = \frac{{ - 1}}{3} + \frac{3}{4}\]

\[2x = \frac{{ - 4}}{{12}} + \frac{9}{{12}}\]

\[2x = \frac{5}{{12}}\]

\[x = \frac{5}{{12}}:2\]

\[x = \frac{5}{{24}}.\]