Đáp án đúng: D
Giải chi tiết:\(4{x^2} + \frac{1}{{{x^2}}} + \left| {2x - \frac{1}{x}} \right| - 6 = 0\,\,\,\,\left( * \right)\)
ĐK: \(x \ne 0\).
Đặt \(t = \left| {2x - \frac{1}{x}} \right|\,\,\left( {t \ge 0} \right)\) ta có: \({t^2} = {\left( {2x - \frac{1}{x}} \right)^2} = 4{x^2} - 4 + \frac{1}{{{x^2}}}\)\( \Rightarrow 4{x^2} + \frac{1}{{{x^2}}} = {t^2} + 4.\)
\(\begin{array}{l} \Rightarrow \left( * \right) \Leftrightarrow {t^2} + 4 + t - 6 = 0 \Leftrightarrow {t^2} + t - 2 = 0\\ \Leftrightarrow \left( {t - 1} \right)\left( {t + 2} \right) = 0 \Leftrightarrow \left[ \begin{array}{l}t - 1 = 0\\t + 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}t = 1\,\,\,\,\left( {tm} \right)\\t = - 2\,\,\,\,\left( {ktm} \right)\end{array} \right.\end{array}\)
\(\begin{array}{l} \Rightarrow \left| {2x - \frac{1}{x}} \right| = 1 \Leftrightarrow \left[ \begin{array}{l}2x - \frac{1}{x} = 1\\2x - \frac{1}{x} = - 1\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2{x^2} - x - 1 = 0\\2{x^2} + x - 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\left( {x - 1} \right)\left( {2x + 1} \right) = 0\\\left( {2x - 1} \right)\left( {x + 1} \right) = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x - 1 = 0\\2x + 1 = 0\\2x - 1 = 0\\x + 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\,\,\,\left( {tm} \right)\\x = - \frac{1}{2}\,\,\,\left( {tm} \right)\\x = \frac{1}{2}\,\,\,\left( {tm} \right)\\x = - 1\,\,\left( {tm} \right)\end{array} \right.\\ \Rightarrow T = 1 - \frac{1}{2} + \frac{1}{2} - 1 = 0.\end{array}\).
Chọn D.
