Ta có \(f'\left( x \right) + \left( {2x - 1} \right){\left[ {f\left( x \right) - x} \right]^2} = 1 \Leftrightarrow f'\left( x \right) - 1 = - \left( {2x - 1} \right){\left[ {f\left( x \right) - x} \right]^2}\)

Đặt \(g\left( x \right) = f\left( x \right) - x \Leftrightarrow g'\left( x \right) = f'\left( x \right) - 1\), do đó \(g'\left( x \right) = - \left( {2x - 1} \right){g^2}\left( x \right)\)

\( \Leftrightarrow - \frac{{g'\left( x \right)}}{{{g^2}\left( x \right)}} = 2x - 1 \Leftrightarrow \int - \frac{{g'\left( x \right)}}{{{g^2}\left( x \right)}}{\rm{d}}x = \int {\left( {2x - 1} \right){\rm{d}}x} \)

\( \Leftrightarrow \frac{1}{{g\left( x \right)}} = {x^2} - x + C \Leftrightarrow \frac{1}{{f\left( x \right) - x}} = {x^2} - x + C\)

Mà \[f\left( 0 \right) = 1\] nên \(\frac{1}{{f\left( 0 \right) - 0}} = C \Leftrightarrow C = 1.\)

Do đó \[f\left( x \right) = x + \frac{1}{{{x^2} - x + 1}} \Rightarrow S = f\left( 1 \right) + f\left( 2 \right) = 2 + \frac{7}{3} = \frac{{13}}{3}.\]

Chọn C.