Tìm x, biết: a) 2/5 + 3/5x = 0

a) \[\frac{2}{5} + \frac{3}{5}x = 0\]

 \[\frac{3}{5}x = - \frac{2}{5}\]

 \[x = \frac{{ - 2}}{5}:\frac{3}{5}\]

 \[x = \frac{{ - 2}}{5} \cdot \frac{5}{3}\]

 \[x = \frac{{ - 2}}{3}.\]

Vậy \[x = \frac{{ - 2}}{3}.\]

b) \(1,3x - 2,5 = - 4\)

 \(1,3x = - 5,1 + 2,5\)

 \(1,3x = - 2,6\)

 \(x = - 2.\)

Vậy \(x = - 2.\)

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:

\[x - 1 = 4\]

\[x = 4 + 1\]

\[x = 5\]

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

Trường hợp 2:

\[x - 1 = - 4\]

\[x = - 4 + 1\]

\[x = - 3\]