Đáp án đúng: D
Giải chi tiết:Vì \(A\left( x \right)\) nhận \(x = \frac{{ - 1}}{2}\) làm nghiệm nên ta có:
\(\begin{array}{l}A\left( { - \frac{1}{2}} \right) = 0\\Hay\,\,\\\left( {2a - 1} \right){\left( { - \frac{1}{2}} \right)^2} - \left( {3 - 4a} \right)\left( { - \frac{1}{2}} \right) + 1 - 6a = 0\\\frac{1}{4}\left( {2a - 1} \right) + \frac{1}{2}\left( {3 - 4a} \right) + 1 - 6a\,\,\,\,\,\,\,\, = 0\\\frac{1}{2}a - \frac{1}{4} + \frac{3}{2} - 2a + 1 - 6a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0\\\left( {\frac{1}{2}a - 2a - 6a} \right) + \left( {\frac{3}{2} - \frac{1}{4} + 1} \right)\,\,\,\,\,\,\,\,\, = 0\\ - \frac{{15}}{2}a + \frac{9}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0\\\frac{{15}}{2}a = \frac{9}{4}\\\,\,\,\,\,\,\,a = \frac{9}{4}.\frac{2}{{15}}\\\,\,\,\,\,\,\,a = \frac{3}{{10}}\end{array}\)
Vậy \(a = \frac{3}{{10}}\)
Chọn D