Đáp án đúng: C
Giải chi tiết:ĐKXĐ: \(8x + 5 \ge 0\)\( \Leftrightarrow x \ge \frac{{ - 5}}{8}\)
Phương trình: \(8{x^2} - 12x - 1 = \sqrt {8x + 5} \)
\(\begin{array}{l} \Leftrightarrow {\left( {4x} \right)^2} - 16x + 4 - 8x - 6 = 2\sqrt {8x + 5} \\ \Leftrightarrow {\left( {4x} \right)^2} - 2.4x.2 + 4 = \left( {8x + 5} \right) + 2\sqrt {8x + 5} + 1\\ \Leftrightarrow {\left( {4x - 2} \right)^2} = {\left( {\sqrt {8x + 5} + 1} \right)^2}\\ \Leftrightarrow \left| {4x - 2} \right| = \sqrt {8x + 5} + 1\\ \Leftrightarrow \left[ \begin{array}{l} - 4x + 2 = \sqrt {8x + 5} + 1\,\,\,\,\left( {voi\,\,\, - \frac{5}{8} \le x < \frac{1}{2}} \right)\\4x - 2 = \sqrt {8x + 5} + 1\,\,\,\,\,\,\,\left( {voi\,\,\,\,x \ge \frac{1}{2}} \right)\end{array} \right.\\ \Leftrightarrow 16{x^2} - 24x - 2 = 2\sqrt {8x + 5} \\ \Leftrightarrow \left[ \begin{array}{l}{\left( { - 4x + 1} \right)^2} = 8x + 5\,\,\,\,\,\left( {voi\,\,\, - \frac{5}{8} \le x < \frac{1}{2}} \right)\\{\left( {4x - 3} \right)^2} = 8x + 5\,\,\,\,\,\,\,\left( {voi\,\,\,\,x \ge \frac{1}{2}} \right)\,\end{array} \right.\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}16{x^2} - 16x - 4 = 0\,\,\,\,\,\left( {voi\,\,\, - \frac{5}{8} \le x < \frac{1}{2}} \right)\\16{x^2} - 32x + 4 = 0\,\,\,\,\,\left( {voi\,\,\,\,x \ge \frac{1}{2}} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{{1 - \sqrt 2 }}{2}\,\\x = \frac{{2 + \sqrt 3 }}{2}\end{array} \right.\end{array}\)
Vậy \(x = \frac{{1 - \sqrt 2 }}{2}\) hoặc \(x = \frac{{2 + \sqrt 3 }}{2}\).
Chọn C.
