Đáp án:
$\begin{array}{l}
a){\left( {x - 1} \right)^2} = 1\\
\Rightarrow {\left( {x - 1} \right)^2} = {1^2} = {\left( { - 1} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 1\\
x - 1 = - 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = 0
\end{array} \right.\\
Vậy\,x = 2\,\,hoặc\,\,x = 0\\
c){\left( {2x - 16} \right)^7} = 128\\
\Rightarrow {\left( {2x - 16} \right)^7} = {2^7}\\
\Rightarrow 2x - 16 = 2\\
\Rightarrow 2x = 18\\
\Rightarrow x = 9\\
Vậy\,x = 9
\end{array}$
