Tìm x biết: x^2 - 8x + 16 = 0

a) \({x^2} - 8x + 16 = 0\)

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

\({\left( {x - 4} \right)^2} = 0\)

Suy ra \(x - 4 = 0\)

\(x = 4\)

Vậy \(x = 4.\)

b) \[4{\left( {x - 1} \right)^2} - \left( {2x + 1} \right)\left( {2x - 1} \right) =  - 3\]

\[4\left( {{x^2} - 2x + 1} \right) - \left( {4{x^2} - 1} \right) =  - 3\]

\[4{x^2} - 8x + 4 - 4{x^2} + 1 =  - 3\]

\[ - 8x + 5 =  - 3\]

\[8x = 8\]

\(x = 1\)

Vậy \(x = 1\).