Đáp án:

\[x = \dfrac{{49}}{9}\]

Giải thích các bước giải:

 ĐKXĐ: \(x \ge \dfrac{5}{2}\)

Ta có:

\(\begin{array}{l}
\sqrt {x + 2 + 3\sqrt {2x - 5} }  + \sqrt {x - 2 - \sqrt {2x - 5} }  + \sqrt {x - \dfrac{1}{2} + 2\sqrt {2x - 5} }  = 5\sqrt 2 \\
 \Leftrightarrow \sqrt 2 .\left( {\sqrt {x + 2 + 3\sqrt {2x - 5} }  + \sqrt {x - 2 - \sqrt {2x - 5} }  + \sqrt {x - \dfrac{1}{2} + 2\sqrt {2x - 5} } } \right) = 5\sqrt 2 .\sqrt 2 \\
 \Leftrightarrow \sqrt {2.\left( {x + 2 + 3\sqrt {2x - 5} } \right)}  + \sqrt {2.\left( {x - 2 - \sqrt {2x - 5} } \right)}  + \sqrt {2.\left( {x - \dfrac{1}{2} + 2\sqrt {2x - 5} } \right)}  = 10\\
 \Leftrightarrow \sqrt {2x + 4 + 6\sqrt {2x - 5} }  + \sqrt {2x - 4 - 2\sqrt {2x - 5} }  + \sqrt {2x - 1 + 4\sqrt {2x - 5} }  = 10\\
 \Leftrightarrow \sqrt {\left( {2x - 5} \right) + 6\sqrt {2x - 5}  + 9}  + \sqrt {\left( {2x - 5} \right) - 2\sqrt {2x - 5}  + 1}  + \sqrt {\left( {2x - 5} \right) + 4\sqrt {2x - 5}  + 4}  = 10\\
 \Leftrightarrow \sqrt {{{\left( {\sqrt {2x - 5}  + 3} \right)}^2}}  + \sqrt {{{\left( {\sqrt {2x - 5}  - 1} \right)}^2}}  + \sqrt {{{\left( {\sqrt {2x - 5}  + 1} \right)}^2}}  = 10\\
 \Leftrightarrow \sqrt {2x - 5}  + 3 + \left| {\sqrt {2x - 5}  - 1} \right| + \sqrt {2x - 5}  + 1 = 10\\
 \Leftrightarrow 2\sqrt {2x - 5}  + \left| {\sqrt {2x - 5}  - 1} \right| = 6\,\,\,\,\,\,\,\,\,\,\,\left( 1 \right)\\
TH1:\,\,\,\,\sqrt {2x - 5}  - 1 > 0 \Leftrightarrow \sqrt {2x - 5}  > 1 \Leftrightarrow x > 3\\
\left( 1 \right) \Leftrightarrow 2\sqrt {2x - 5}  + \left( {\sqrt {2x - 5}  - 1} \right) = 6\\
 \Leftrightarrow 3\sqrt {2x - 5}  = 7\\
 \Leftrightarrow \sqrt {2x - 5}  = \dfrac{7}{3}\\
 \Leftrightarrow 2x - 5 = \dfrac{{49}}{9}\\
 \Leftrightarrow x = \dfrac{{47}}{9}\,\,\,\,\,\,\,\,\,\,\left( {t/m} \right)\\
TH2:\,\,\,\,\,\,\,\,\sqrt {2x - 5}  - 1 \le 0 \Leftrightarrow \sqrt {2x - 5}  \le 1 \Leftrightarrow \dfrac{5}{2} \le x \le 3\\
\left( 1 \right) \Leftrightarrow 2\sqrt {2x - 5}  + \left( {1 - \sqrt {2x - 5} } \right) = 6\\
 \Leftrightarrow \sqrt {2x - 5}  = 5\,\,\,\,\,\,\,\,\,\,\,\left( {L,\,\,\,\sqrt {2x - 5}  \le 1} \right)
\end{array}\)

Vậy \(x = \dfrac{{49}}{9}\)