Đáp án đúng: A
Giải chi tiết:Đặt \(\dfrac{{f\left( x \right) - 3}}{{2x - 1}} = g\left( x \right) \Rightarrow f\left( x \right) = \left( {2x - 1} \right)g\left( x \right) + 3\)
\( \Rightarrow \mathop {\lim }\limits_{x \to \dfrac{1}{2}} f\left( x \right) = 3\).
\(\begin{array}{l}L = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{\sqrt {2f\left( x \right) + 3}  + \sqrt {f\left( x \right) + 1}  - 5}}{{2{x^2} - x}}\\\,\,\,\, = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{\sqrt {2f\left( x \right) + 3}  - 3 + \sqrt {f\left( x \right) + 1}  - 2}}{{2{x^2} - x}}\\\,\,\,\, = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{\sqrt {2f\left( x \right) + 3}  - 3}}{{2{x^2} - x}} + \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{\sqrt {f\left( x \right) + 1}  - 2}}{{2{x^2} - x}}\\\,\,\,\, = \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{2f\left( x \right) + 3 - 9}}{{\left( {2{x^2} - x} \right)\left[ {\sqrt {2f\left( x \right) + 3}  + 3} \right]}} + \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{f\left( x \right) + 1 - 4}}{{\left( {2{x^2} - x} \right)\left[ {\sqrt {f\left( x \right) + 1}  + 2} \right]}}\\\,\,\,\, = 2\mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{f\left( x \right) - 3}}{{\left( {2x - 1} \right)}}.\dfrac{1}{{x\left[ {\sqrt {2f\left( x \right) + 3}  + 3} \right]}} + \mathop {\lim }\limits_{x \to \dfrac{1}{2}} \dfrac{{f\left( x \right) - 3}}{{\left( {2x - 1} \right)}}.\dfrac{1}{{x\left[ {\sqrt {f\left( x \right) + 1}  + 2} \right]}}\\\,\,\,\, = 2.5.\dfrac{1}{{\dfrac{1}{2}.\left( {\sqrt {2.3 + 3}  + 3} \right)}} + 5.\dfrac{1}{{\dfrac{1}{2}\left( {\sqrt {3 + 1}  + 2} \right)}}\\\,\,\,\, = \dfrac{{10}}{3} + \dfrac{5}{2} = \dfrac{{35}}{6}\\ \Rightarrow a = 35,\,\,b = 6 \Rightarrow a + b = 41.\end{array}\)
Chọn A.