Tính 2b – a.

Ta có \(f'\left( x \right) = \frac{{x + 1}}{{2018x}}.{\left( {\frac{{2018x}}{{x + 1}}} \right)^\prime } = \frac{1}{{x\left( {x + 1} \right)}} = \frac{1}{x} - \frac{1}{{x + 1}}\).

Do đó \(f'\left( 1 \right) = \frac{1}{1} - \frac{1}{2};f'\left( 2 \right) = \frac{1}{2} - \frac{1}{3};...;f'\left( {2018} \right) = \frac{1}{{2018}} - \frac{1}{{2019}}\).

Suy ra S = f'(1) + f'(2) + …+f'(2018) = \(1 - \frac{1}{{2019}} = \frac{{2018}}{{2019}}\).

Suy ra a = 2018; b = 2019. Do đó 2b – a = 2020.

Trả lời: 2020.