Ta có \(\left( {x - 1} \right)f'\left( {\rm{x}} \right) = \frac{{f\left( {\rm{x}} \right)}}{{x + 2}} \Leftrightarrow \frac{{f'\left( {\rm{x}} \right)}}{{f\left( {\rm{x}} \right)}} = \frac{1}{{\left( {x - 1} \right)\left( {x + 2} \right)}}\).                      

Lấy nguyên hàm hai vế ta có \[\int {\frac{{f'\left( {\rm{x}} \right)}}{{f\left( {\rm{x}} \right)}}dx} = \int {\frac{{dx}}{{\left( {x - 1} \right)\left( {x + 2} \right)}}} \] suy ra \[\ln \left| {f\left( {\rm{x}} \right)} \right| = \frac{1}{3}\ln \left| {\frac{{x - 1}}{{x + 2}}} \right| + C\]

Do \(f(2) = 2\) nên \(\ln 2 = \frac{1}{3}\ln \frac{1}{4} + C \Leftrightarrow C = \frac{{5\ln 2}}{3} = \frac{{\ln 32}}{3}\).

Suy ra \(\ln \left| {f\left( {\rm{x}} \right)} \right| = \frac{1}{3}\left( {\ln \left| {\frac{{x - 1}}{{x + 2}}} \right| + \ln 32} \right) = \ln \left( {\sqrt[3]{{32 \cdot \frac{{x - 1}}{{x + 2}}}}} \right)\).

Vậy \(\left| {f\left( {\rm{x}} \right)} \right| = \sqrt[3]{{32 \cdot \frac{{x - 1}}{{x + 2}}}} \cdot \) Ta có \(\left| {f\left( {\frac{{86}}{{85}}} \right)} \right| = \frac{1}{2}\). Chọn D.