Biết phương trình \({\log _5}\frac{{2\sqrt x  + 1}}{x} = 2{\log _3}\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right)\) có

ĐKXĐ: x > 1

Ta có:

\(\begin{array}{l}
{\log _5}\frac{{2\sqrt x  + 1}}{x} = 2{\log _3}\left( {\frac{{\sqrt x }}{2} - \frac{1}{{2\sqrt x }}} \right) \Leftrightarrow {\log _5}\frac{{2\sqrt x  + 1}}{x} = 2{\log _3}\left( {\frac{{x - 1}}{{2\sqrt x }}} \right)\\
 \Leftrightarrow {\log _5}\left( {2\sqrt x  + 1} \right) - {\log _5}x = 2{\log _3}\left( {x - 1} \right) - 2{\log _3}\left( {2\sqrt x } \right)\\
 \Leftrightarrow {\log _5}\left( {2\sqrt x  + 1} \right) + 2{\log _3}\left( {2\sqrt x } \right) = {\log _5}x + 2{\log _3}\left( {x - 1} \right)\,\,\,\,(1)
\end{array}\)

Xét hàm số \(f\left( t \right) = {\log _5}t + 2{\log _3}\left( {t - 1} \right),t \in \left( {1; + \infty } \right)\) , có:  \(f'\left( t \right) = \frac{1}{{t.\ln 5}} + \frac{2}{{\left( {t - 1} \right).\ln 3}} > 0,\forall t \in (1; + \infty )\)

=> Hàm số f(t) đồng biến trên \(\left( {1; + \infty } \right)\)

Khi đó, phương trình \(\left( 1 \right) \Leftrightarrow f\left( {2\sqrt x  + 1} \right) = f\left( x \right) \Leftrightarrow 2\sqrt x  + 1 = x \Leftrightarrow x - 2\sqrt x  - 1 = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}
{\sqrt x  = 1 + \sqrt 2 }\\
{\sqrt x  = 1 - \sqrt 2  < 0}
\end{array}} \right.\)

\( \Leftrightarrow \sqrt x  = 1 + \sqrt 2  \Leftrightarrow x = {\left( {1 + \sqrt 2 } \right)^2} = 3 + 2\sqrt 2  \Rightarrow a = 3,b = 2 =  > 2a + b = 2.3 + 2 = 8\)