Phương pháp giải:

Chuyển vế, lấy nguyên hàm hai vế.

Lời giải chi tiết:

\(\begin{array}{l}
\,\,\,\,\,f'\left( x \right) = \left( {2x + 1} \right){f^2}\left( x \right)\\
\Leftrightarrow \frac{{f'\left( x \right)}}{{{f^2}\left( x \right)}} = 2x + 1\\
\Leftrightarrow \int\limits_{}^{} {\frac{{f'\left( x \right)dx}}{{{f^2}\left( x \right)}}} = \int\limits_{}^{} {\left( {2x + 1} \right)dx} \\
\Leftrightarrow \frac{{ - 1}}{{f\left( x \right)}} = {x^2} + x + C\\
f\left( 1 \right) = - 0,5 \Leftrightarrow - \frac{1}{{ - 0,5}} = 1 + 1 + C \Leftrightarrow C = 0\\
\Leftrightarrow f\left( x \right) = - \frac{1}{{{x^2} + x}} = - \frac{1}{{x\left( {x + 1} \right)}} = - \left( {\frac{1}{x} - \frac{1}{{x + 1}}} \right) = \frac{1}{{x + 1}} - \frac{1}{x}\\
\Rightarrow f\left( 1 \right) + f\left( 2 \right) + f\left( 3 \right) + ... + f\left( {2017} \right)\\
= \frac{1}{2} - 1 + \frac{1}{3} - \frac{1}{2} + \frac{1}{4} - \frac{1}{3} + ... + \frac{1}{{2017}} - \frac{1}{{2016}} + \frac{1}{{2018}} - \frac{1}{{2017}}\\
= - 1 + \frac{1}{{2018}} = \frac{{ - 2017}}{{2018}} = \frac{a}{b} \Rightarrow \left\{ \begin{array}{l}
a = - 2017\\
b = 2018
\end{array} \right. \Rightarrow b - a = 4035
\end{array}\)

Chọn B.