Tìm \(x \in \mathbb{Z},\) biết:

g) \(\frac{{x + 1}}{6} = \frac{2}{x}.\)

h) \(\frac{4}{{15}}x + \frac{2}{3} =  - \frac{1}{5}.\)

\(\frac{4}{{15}}x =  - \frac{1}{5} - \frac{2}{3}\)

\[\frac{4}{{15}}x =  - \frac{3}{{15}} - \frac{{10}}{{15}}\]

\[\frac{4}{{15}}x = \frac{{ - 13}}{{15}}\]

\[x = \frac{{ - 13}}{{15}}:\frac{4}{{15}}\]

\[x = \frac{{ - 13}}{{15}} \cdot \frac{{15}}{4}\]

\[x = \frac{{ - 13}}{4}.\]

k) \(\frac{3}{4}:x - \frac{1}{3} = \frac{5}{{14}} \cdot \frac{{ - 7}}{6}\)

\(\frac{3}{4}:x - \frac{1}{3} = \frac{{5 \cdot \left( { - 7} \right)}}{{14 \cdot 6}}\)

\(\frac{3}{4}:x - \frac{1}{3} = \frac{{ - 5}}{{12}}\)

\(\frac{3}{4}:x = \frac{{ - 5}}{{12}} + \frac{1}{3}\)

\(\frac{3}{4}:x = \frac{{ - 5}}{{12}} + \frac{4}{{12}}\)

\(\frac{3}{4}:x = \frac{{ - 1}}{{12}}\)

\(x = \frac{3}{4}:\frac{{ - 1}}{{12}}\)

\(x = \frac{3}{4} \cdot \left( { - 12} \right)\)

\(x =  - 9.\)