Tìm x, biết:

    a) \(x - 0,75 =  - 0,6\);             b) \( - x - \frac{3}{2} = \frac{{ - 5}}{4}\);    c) \(\frac{1}{9} - {\left( {2x + \frac{1}{2}} \right)^2} = 0\).

a) \(x - 0,75 =  - 0,6\)

\(x =  - 0,6 + 0,75\)

\(x = 0,15\)

Vậy \(x = 0,15\).

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

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

\( - x = \frac{1}{4}\)

\(x =  - \frac{1}{4}\)

Vậy \(x =  - \frac{1}{4}\).

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

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

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:

\(2x + \frac{1}{2} =  - \frac{1}{3}\)

\(2x =  - \frac{1}{3} - \frac{1}{2}\)

\(2x =  - \frac{5}{6}\)

\(x =  - \frac{5}{{12}}\)