Tìm x , biết: 1/ 9 − ( 2 x + 1/ 2 )^ 2 = 0.

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

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

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:

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

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

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

\(x = - \frac{5}{{12}}\).