Đáp án đúng: B
Giải chi tiết:\(\begin{array}{l}\,\,\,\,{\left( {x - 1} \right)^2} + \frac{3}{{12}} = \sqrt {\frac{1}{{16}}} \\\,\,\,\,\,{\left( {x - 1} \right)^2} + \frac{1}{4} = \frac{1}{4}\\\,\,\,\,\,{\left( {x - 1} \right)^2}\,\,\,\,\,\,\,\,\,\, = 0\\\,\,\,\,\,\,\,x - 1\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0\\ \Rightarrow \,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1\end{array}\)
Chọn B