Tìm \(x\), biết:

    a) \(0,75 - x =  - 0,6\);                                 b) \(\frac{3}{4} + x =  - \frac{1}{2}\);

    c) \[\frac{2}{3}:x = 1,4 - \frac{{12}}{5}\]; d) \(\frac{1}{9} - {\left( {2x + \frac{1}{2}} \right)^2} = 0\).

a) \(0,75 - x =  - 0,6\)

\(x = 0,75 - \left( { - 0,6} \right)\)

\(x = 0,75 + 0,6\)

\(x = 1,35\)

Vậy \(x = 1,35\).

c) \[\frac{2}{3}:x = 1,4 - \frac{{12}}{5}\]

\[\frac{2}{3}:x = \frac{{14}}{{10}} - \frac{{12}}{5}\]

\[\frac{2}{3}:x = \frac{7}{5} - \frac{{12}}{5}\]

\[\frac{2}{3}:x =  - 1\]

\[x = \frac{2}{3}:\left( { - 1} \right)\]

\[x =  - \frac{2}{3}\]

Vậy \[x =  - \frac{2}{3}\].

b) \(\frac{3}{4} + x =  - \frac{1}{2}\)

\(x =  - \frac{1}{2} - \frac{3}{4}\)

\(x = \frac{{ - 5}}{4}\)

Vậy \(x = \frac{{ - 5}}{4}\).

d) \(\frac{1}{9} - {\left( {2x + \frac{1}{2}} \right)^2} = 0\)

    \({\left( {2x + \frac{1}{2}} \right)^2} = \frac{1}{9}\)

Trường hợp 1:

\(2x + \frac{1}{2} = \frac{1}{3}\)

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

\(2x =  - \frac{1}{6}\)

\(x =  - \frac{1}{{12}}\)

Vậy \(x \in \left\{ { - \frac{1}{{12}}; - \frac{5}{{12}}} \right\}\).

Trường hợp 2:

\(2x + \frac{1}{2} =  - \frac{1}{3}\)

\(2x =  - \frac{1}{3} - \frac{1}{2}\)

\(2x =  - \frac{5}{6}\)

\(x =  - \frac{5}{{12}}\)