a) \(2 - 25x^2= 0 \)
\( (\sqrt2)^2 - (5x)^2 = 0\)
\( (\sqrt 2 - 5x)( \sqrt 2 + 5x) = 0\)
\( \Rightarrow \left[ \begin{array}{l}\sqrt 2 - 5{\rm{x}} = 0\\\sqrt 2 + 5{\rm{x}} = 0\end{array} \right. \Rightarrow \left[ \begin{array}{l}x = \dfrac{{\sqrt 2 }}{5}\\x = \dfrac{{ - \sqrt 2 }}{5}\end{array} \right.\)
Vậy \(x = \dfrac{{\sqrt 2 }}{5}\) hoặc \(x = \dfrac{{ - \sqrt 2 }}{5}\)
b) \(x^2- x + \dfrac{1}{4} = 0\)
\( x^2- 2 . x . \dfrac{1}{2} + {\left( {\dfrac{1}{2}} \right)^2}= 0\)
\({\left( {x - \dfrac{1}{2}} \right)^2} = 0 \)
\( \Rightarrow x - \dfrac{1}{2}= 0 \Rightarrow x = \dfrac{1}{2}\)
Vậy \(x = \dfrac{1}{2}.\)
