Tìm (x) biết:

a) \(16{x^2} - 16x + 4 = 0\)

\(4{x^2} - 4x + 1 = 0\)

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

\(2x - 1 = 0\)

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

Vậy \(x = \frac{1}{2}\).

b) \[{\left( {x + 3} \right)^2} + \left( {5 - x} \right)\left( {5 + x} \right) =  - 3\]

\[{x^2} + 6x + 9 + 25 - {x^2} =  - 3\]

\[6x + 9 + 25 =  - 3\]

\[6x =  - 37\]

\(x = \frac{{ - 37}}{6}\)

Vậy \(x = \frac{{ - 37}}{6}\).