Cho hàm số \({\rm{f}}\left( {\rm{x}} \right)\) có \({\rm{f}}(2) = 0\) và \({\rm{f'}}\left( {\rm{x}} \right) = \frac{{{\rm{x}} + 7}}{{\sqrt {2{\rm{x}} - 3} }},\,\,\forall {\rm{x}} \in \left( {\frac{3}{2}\,;\, + \infty } \right)\). Biết rằng \(\int\limits_4^7 {f\left( {\frac{{\rm{x}}}{2}} \right){\rm{dx}} = \frac{{\rm{a}}}{{\rm{b}}}} \) \(\left( {{\rm{a}},\,\,{\rm{b}} \in \mathbb{Z}\,,\,\,{\rm{b}} > 0\,,\,\,\frac{{\rm{a}}}{{\rm{b}}}} \right.\) là phân số tối giản). Khi đó \({\rm{a}} + {\rm{b}}\) ...

Ta có: \[\begin{array}{l}\int\limits_4^{\frac{7}{2}} {f\left( {\frac{x}{2}} \right)dx}  = 2\int\limits_2^{\frac{7}{2}} {f\left( {\frac{x}{2}} \right)d\left( {\frac{x}{2}} \right)}  = 2\int\limits_2^{\frac{7}{2}} {f\left( x \right)dx}  = 2\int\limits_2^{\frac{7}{2}} {f\left( x \right)d\left( {x - \frac{7}{2}} \right)} \\\end{array}\] .

Đặt \(\left\{ \begin{array}{l}u = f\left( x \right)\\dv = d\left( {x - \frac{7}{2}} \right)\end{array} \right. \Rightarrow \left\{ \begin{array}{l}du = f'\left( x \right)dx\\v = \left( {x - \frac{7}{2}} \right)\end{array} \right.\) .

Khi đó: \[2\int\limits_2^{\frac{7}{2}} {f\left( x \right)dx}  = 2\int\limits_2^{\frac{7}{2}} {f\left( x \right)d\left( {x - \frac{7}{2}} \right)}  = 2\left[ {\left. {\left( {x - \frac{7}{2}} \right)f\left( x \right)} \right|_0^{\frac{7}{2}} - \int\limits_2^{\frac{7}{2}} {\left( {x - \frac{7}{2}} \right)f'\left( x \right)dx} } \right]\]

\( =  - 2\int\limits_2^{\frac{7}{2}} {\left( {x - \frac{7}{2}} \right)f'\left( x \right)dx}  =  - 2\int\limits_2^{\frac{7}{2}} {\left( {x - \frac{7}{2}} \right)\frac{{\sqrt {2x - 3} }}dx}  = \frac.\)

Do đó \(a = 236\,;\,\,b = 15 \Rightarrow a + b = 251.\) Chọn B.