Đáp án đúng: D

Giải chi tiết:\(11\sqrt {5 - x}  + 8\sqrt {2x - 1}  = 24 + 3\sqrt {\left( {5 - x} \right)\left( {2x - 1} \right)} \,\,\,\left( * \right)\)
ĐKXĐ: \(\left\{ \begin{array}{l}5 - x \ge 0\\2x - 1 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \le 5\\x \ge \dfrac{1}{2}\end{array} \right. \Leftrightarrow \dfrac{1}{2} \le x \le 5.\)
Đặt \(\left\{ \begin{array}{l}\sqrt {5 - x}  = a\,\,\,\left( {a \ge 0} \right)\\\sqrt {2x - 1}  = b\,\,\,\left( {b \ge 0} \right)\end{array} \right.\)\( \Rightarrow \left\{ \begin{array}{l}{a^2} = 5 - x\\{b^2} = 2x - 1\end{array} \right.\)
\( \Rightarrow 2{a^2} + {b^2} = 2\left( {5 - x} \right) + 2x - 1 = 9.\) 
 Khi đó ta có: \(\left\{ \begin{array}{l}11a + 8b = 24 + 3ab\,\,\,\,\,\left( 1 \right)\\2{a^2} + {b^2} = 9\,\,\,\,\,\left( 2 \right)\end{array} \right.\)
+) Giải phương trình \(\left( 1 \right)\) ta có:
\(\left( 1 \right) \Leftrightarrow 11a - 3ab = 24 - 8b\) \( \Leftrightarrow a\left( {11 - 3b} \right) = 24 - 8b\,\,\,\left( * \right)\)
Với \(11 - 3b = 0 \Leftrightarrow b = \dfrac{{11}}{3}\) \( \Rightarrow \left( * \right) \Leftrightarrow 0a =  - \dfrac{{16}}{3}\) (vô lý)
\( \Rightarrow b = \dfrac{{11}}{3}\) không là nghiệm của phương trình \(\left( * \right).\)
\( \Rightarrow \left( * \right) \Leftrightarrow a = \dfrac{{24 - 8b}}{{11 - 3b}} = \dfrac{{8b - 24}}{{3b - 11}}\)
Thay \(a = \dfrac{{8b - 24}}{{3b - 11}}\) vào \(\left( 2 \right)\) ta được:
\(\begin{array}{l}\left( 2 \right) \Leftrightarrow 2{\left( {\dfrac{{8b - 24}}{{3b - 11}}} \right)^2} + {b^2} = 9\\ \Leftrightarrow 2\left( {64{b^2} - 384b + 576} \right) + {b^2}\left( {9{b^2} - 66b + 121} \right) = 9\left( {9{b^2} - 66b + 121} \right)\\ \Leftrightarrow 128{b^2} - 768b + 1152 + 9{b^4} - 66{b^3} + 121{b^2} - 81{b^2} + 594b - 1089 = 0\\ \Leftrightarrow 9{b^4} - 66{b^3} + 168{b^2} - 174b + 63 = 0\\ \Leftrightarrow 3{b^4} - 22{b^3} + 56{b^2} - 58b + 21 = 0\\ \Leftrightarrow \left( {b - 1} \right)\left( {3{b^3} - 19{b^2} + 37b - 21} \right) = 0\\ \Leftrightarrow \left( {b - 1} \right)\left( {b - 1} \right)\left( {b - 3} \right)\left( {3b - 7} \right) = 0\end{array}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}b - 1 = 9\\b - 3 = 0\\3b - 7 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}b = 1\,\,\,\left( {tm} \right)\\b = 3\,\,\,\,\left( {tm} \right)\\b = \dfrac{7}{3}\,\,\left( {tm} \right)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\sqrt {2x - 1}  = 1\\\sqrt {2x - 1}  = 3\\\sqrt {2x - 1}  = \dfrac{7}{3}\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}2x - 1 = 1\\2x - 1 = 9\\2x - 1 = \dfrac{{49}}{9}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}2x = 2\\2x = 10\\2x = \dfrac{{58}}{9}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\,\,\,\left( {tm} \right)\\x = 5\,\,\left( {tm} \right)\\x = \dfrac{{29}}{9}\,\,\,\left( {tm} \right)\end{array} \right.\end{array}\)
Vậy phương trình có tập nghiệm:\(S = \left\{ {1;\,\dfrac{{29}}{9};\,\,5} \right\}.\)
Chọn D.