Tìm \(x \in \mathbb{Z},\) biết:

a) \(\frac{9}{x} = \frac{{ - 35}}{{105}}.\)    

b) \(\frac{6}{{ - x}} = \frac{x}{{ - 24}}.\)

c) \(\frac{{x - 2}}{5} = \frac{{1 - x}}{6}.\)   

d) \(\frac{{x - 3}}{{ - 2}} = \frac{{ - 8}}{{x - 3}}.\)

e) \(\frac{{ - 2}}{5} < \frac{x}{{15}} < \frac{1}{6}.\)     

f) \(\frac{{ - 1}}{4}:\frac{{ - 3}}{4} + \frac{1}{2} < x < \frac{7}{8} - \frac{1}{2}:\frac{{ - 5}}{6}.\)

a) \(\frac{9}{x} = \frac{{ - 35}}{{105}}\)

\(\begin{array}{l}x = \frac{{105 \cdot 9}}{{ - 35}}\\x =  - 27\end{array}\)

Vậy \(x =  - 27.\)

c) \(\frac{{x - 2}}{5} = \frac{{1 - x}}{6}\)

\(6\left( {x - 2} \right) = 5\left( {1 - x} \right)\)

\(6x - 12 = 5 - 5x\)

\(6x + 5x = 5 + 12\)

\(11x = 17\)

\(x = \frac{{17}}{{11}}.\)

Vậy \(x = \frac{{17}}{{11}}.\)

e) \(\frac{{ - 2}}{5} < \frac{x}{{15}} < \frac{1}{6}\)

\(\frac{{ - 12}}{{30}} < \frac{{2x}}{{30}} < \frac{5}{{30}}\)

Suy ra \( - 12 < 2x < 5\)  

Hay \( - 6 < x < \frac{5}{2}\)   

Vì \(x \in \mathbb{Z}\) nên:

\[x \in \left\{ { - 5;\,\, - 4;\,\, - 3;\,\, - 2;\,\, - 1;\,\,0;\,\,1;\,\,2} \right\}.\]

Vậy \[x \in \left\{ { - 5;\,\, - 4;\,\, - 3;\,\, - 2;\,\, - 1;\,\,0;\,\,1;\,\,2} \right\}.\]

b) \(\frac{6}{{ - x}} = \frac{x}{{ - 24}}\)

\( - {x^2} =  - 144\)

  \({x^2} = 144\)

  \(x = 12\) hoặc \(x =  - 12\)

Vậy \(x \in \left\{ {12;\,\, - 12} \right\}.\)

d) \(\frac{{x - 3}}{{ - 2}} = \frac{{ - 8}}{{x - 3}}\)

\({\left( {x - 3} \right)^2} = 16 = {4^2} = {\left( { - 4} \right)^2}\)

\(x - 3 = 4\) hoặc \(x - 3 =  - 4\)

     \(x = 7\) hoặc \(x =  - 1.\)

Vậy \(x \in \left\{ {7;\,\, - 1} \right\}.\)

f) \(\frac{{ - 1}}{4}:\frac{{ - 3}}{4} + \frac{1}{2} < x < \frac{7}{8} - \frac{1}{2}:\frac{{ - 5}}{6}\)

\(\frac{{ - 1}}{4} \cdot \frac{{ - 4}}{3} + \frac{1}{2} < x < \frac{7}{8} - \frac{1}{2} \cdot \frac{{ - 6}}{5}\)

\(\frac{1}{3} + \frac{1}{2} < x < \frac{7}{8} + \frac{3}{5}\)

\(\frac{2}{6} + \frac{3}{6} < x < \frac{{35}}{{40}} + \frac{{24}}{{40}}\)

\(\frac{5}{6} < x < \frac{{59}}{{40}}\)

Mà \(x \in \mathbb{Z}\) nên \(x = 1.\)

Vậy \(x = 1.\)