\({\rm{a}})\;{\left( {x - \displaystyle {1 \over 2}} \right)^2} = 0 \)
\(\Rightarrow x - \displaystyle {1 \over 2} = 0 \)
\(\Rightarrow x = \displaystyle {1 \over 2}\)
\(b)\;{\left( {x - 2} \right)^2} = 1 \)
\( \Rightarrow \left( {x - 2} \right) = {1^2}\)
\(\Rightarrow \left[ \matrix{
x - 2 = 1 \hfill \cr
x - 2 = - 1 \hfill \cr} \right.\)
\(\Rightarrow \left[ \matrix{
x = 3 \hfill \cr
x = 1 \hfill \cr} \right.\)
\(c)\;{\left( {2{\rm{x}} - 1} \right)^3} = - 8\)
\(\Rightarrow {\left( {2{\rm{x}} - 1} \right)^3} = {\left( -2 \right)^3}\)
\(\Rightarrow 2{\rm{x}} - 1 = - 2\)
\( \Rightarrow 2x = - 2 + 1\)
\( \Rightarrow 2x = - 1\)
\(\Rightarrow x = \displaystyle - {1 \over 2}\)
\(\displaystyle {\rm{d)}}\;{\left( {x + {1 \over 2}} \right)^2} = {1 \over {16}}\)
\(\displaystyle \Rightarrow {\left( {x + {1 \over 2}} \right)^2} = {\left( {{1 \over 4}} \right)^2} \)
\( \Rightarrow \left[ \begin{array}{l}x + \dfrac{1}{2} = \dfrac{1}{4}\\x + \dfrac{1}{2} = - \dfrac{1}{4}\end{array} \right.\)
\( \Rightarrow \left[ \begin{array}{l}x = \dfrac{1}{4} - \dfrac{1}{2}\\x = - \dfrac{1}{4} - \dfrac{1}{2}\end{array} \right.\)
\( \Rightarrow \left[ \begin{array}{l}x = - \dfrac{1}{4}\\x = - \dfrac{3}{4}\end{array} \right.\)
