Tìm \(x,\) biết: a) \(x - \frac{5}{6} = \frac{{ - 7}}{6}.\)                             b) \[x + 1,05 = 0,2 - 4,25.\]         c) \[\frac{{x - 1}}{2} = \frac{8}{{x - 1}}.\]

a) \(x - \frac{5}{6} = \frac{{ - 7}}{6}\)

\(x = \frac{{ - 7}}{6} + \frac{5}{6}\)

\(x = \frac{{ - 2}}{6}\)

\(x = \frac{{ - 1}}{3}.\)

Vậy \(x = \frac{{ - 1}}{3}.\)

b) \[x + 1,05 = 0,2 - 4,25\]

\[x + 1,05 =  - 4,05\]

\[x =  - 4,05 - 1,05\]

\[x =  - 5,1\].

Vậy \[x =  - 5,1.\]

c) \[\frac{{x - 1}}{2} = \frac{8}{{x - 1}}\]

 \[{\left( {x - 1} \right)^2} = 16\]

 \[{\left( {x - 1} \right)^2} = {4^2} = {\left( { - 4} \right)^2}\]

Trường hợp 1:

\[x - 1 = 4\]

\[x = 4 + 1\]

\[x = 5\]

Trường hợp 2:

\[x - 1 =  - 4\]

\[x =  - 4 + 1\]

\[x =  - 3\]Vậy \(x \in \left\{ {5;\,\, - 3} \right\}.\)