Đáp án:

$\begin{array}{l}
a)P = \dfrac{{\dfrac{1}{{2018}} + \dfrac{1}{{2019}} - \dfrac{1}{{2020}}}}{{\dfrac{5}{{2018}} + \dfrac{5}{{2019}} - \dfrac{5}{{2020}}}} - \dfrac{{\dfrac{{ - 2}}{{2017}} - \dfrac{2}{{2018}} + \dfrac{2}{{2019}}}}{{\dfrac{3}{{2017}} + \dfrac{3}{{2018}} - \dfrac{3}{{2019}}}}\\
 = \dfrac{{\dfrac{1}{{2018}} + \dfrac{1}{{2019}} - \dfrac{1}{{2020}}}}{{5.\left( {\dfrac{1}{{2018}} + \dfrac{1}{{2019}} - \dfrac{1}{{2020}}} \right)}} - \dfrac{{ - 2.\left( {\dfrac{1}{{2017}} + \dfrac{1}{{2018}} - \dfrac{1}{{2019}}} \right)}}{{3.\left( {\dfrac{1}{{2017}} + \dfrac{1}{{2018}} - \dfrac{1}{{2019}}} \right)}}\\
 = \dfrac{1}{5} - \dfrac{{ - 2}}{3}\\
 = \dfrac{1}{5} + \dfrac{2}{3}\\
 = \dfrac{{3 + 10}}{{15}} = \dfrac{{13}}{{15}}\\
b)\\
A = \left( {\dfrac{{1,5 + 1 - 0,75}}{{2,5 + \dfrac{5}{3} - 1,25}} + \dfrac{{0,375 - 0,3 + \dfrac{3}{{11}} + \dfrac{3}{{12}}}}{{ - 0,625 + 0,5 - \dfrac{5}{{11}} - \dfrac{5}{{12}}}}} \right)\\
:\dfrac{{1890}}{{2005}} + 115\\
 = \left( \begin{array}{l}
\dfrac{{\left( {1,5 + 1 - 0,75} \right).\dfrac{5}{3}.\dfrac{3}{5}}}{{2,5 + \dfrac{5}{3} - 1,25}}\\
 - \dfrac{{\left( {0,375 - 0,3 + \dfrac{3}{{11}} + \dfrac{3}{{12}}} \right).\dfrac{5}{3}.\dfrac{3}{5}}}{{0,625 - 0,5 + \dfrac{5}{{11}} + \dfrac{5}{{12}}}}
\end{array} \right):\dfrac{{1890}}{{2005}} + 115\\
 = \left( {\dfrac{{\left( {2,5 + \dfrac{5}{3} - 1,25} \right).\dfrac{3}{5}}}{{2,5 + \dfrac{5}{3} - 1,25}} - \dfrac{{\left( {0,625 - 0,5 + \dfrac{5}{{11}} + \dfrac{5}{{12}}} \right).\dfrac{3}{5}}}{{0,625 - 0,5 + \dfrac{5}{{11}} + \dfrac{5}{{12}}}}} \right)\\
:\dfrac{{1890}}{{2005}} + 115\\
 = \left( {\dfrac{3}{5} - \dfrac{3}{5}} \right):\dfrac{{1890}}{{2005}} + 115\\
 = 0:\dfrac{{1890}}{{2005}} + 115\\
 = 115\\
c)C = \dfrac{{\dfrac{1}{6} - \dfrac{1}{{39}} + \dfrac{1}{{51}}}}{{\dfrac{1}{8} - \dfrac{1}{{52}} + \dfrac{1}{{68}}}}\\
 = \dfrac{{\left( {\dfrac{1}{6} - \dfrac{1}{{39}} + \dfrac{1}{{51}}} \right).\dfrac{3}{4}.\dfrac{4}{3}}}{{\dfrac{1}{8} - \dfrac{1}{{52}} + \dfrac{1}{{68}}}}\\
 = \dfrac{{\left( {\dfrac{1}{8} - \dfrac{1}{{52}} + \dfrac{1}{{68}}} \right).\dfrac{4}{3}}}{{\dfrac{1}{8} - \dfrac{1}{{52}} + \dfrac{1}{{68}}}}\\
 = \dfrac{4}{3}
\end{array}$