Đáp án:

 C5: \(\dfrac{{13}}{{15}}\)

Giải thích các bước giải:

\(\begin{array}{l}
C1:\\
3 - \dfrac{1}{4} + \dfrac{2}{3} - 5 - \dfrac{1}{3} + \dfrac{6}{5} - 6 + \dfrac{7}{4} - \dfrac{3}{2}\\
 = \left( {3 - 5 - 6} \right) + \left( { - \dfrac{1}{4} - \dfrac{3}{2} + \dfrac{7}{4}} \right) + \left( {\dfrac{2}{3} - \dfrac{1}{3}} \right) + \dfrac{6}{5}\\
 =  - 8 + \left( {\dfrac{{ - 1 - 6 + 7}}{4}} \right) + \left( {\dfrac{{2 - 1}}{3}} \right) + \dfrac{6}{5}\\
 =  - 8 + \dfrac{1}{3} + \dfrac{6}{5}\\
 = \dfrac{{ - 8.15 + 5 + 18}}{{15}} = \dfrac{{ - 120 + 23}}{{15}} =  - \dfrac{{97}}{{15}}\\
C2:8 - \dfrac{9}{4} + \dfrac{2}{7} + 6 + \dfrac{3}{7} - \dfrac{5}{4} - 3 - \dfrac{2}{4} + \dfrac{9}{7}\\
 = \left( {8 + 6 - 3} \right) + \left( { - \dfrac{9}{4} - \dfrac{5}{4} - \dfrac{2}{4}} \right) + \left( {\dfrac{2}{7} + \dfrac{3}{7} + \dfrac{9}{7}} \right)\\
 = 11 + \dfrac{{ - 9 - 5 - 2}}{4} + \dfrac{{14}}{7} = 11 - 4 + 2 = 9\\
C3:\\
 - \dfrac{1}{2} + \dfrac{3}{5} - \dfrac{1}{9} + \dfrac{1}{{131}} + \dfrac{2}{7} + \dfrac{4}{{35}} - \dfrac{7}{{18}}\\
 = \left( { - \dfrac{1}{2} - \dfrac{1}{9} - \dfrac{7}{{18}}} \right) + \left( {\dfrac{3}{5} + \dfrac{2}{7} + \dfrac{4}{{35}}} \right) + \dfrac{1}{{131}}\\
 = \left( {\dfrac{{ - 9 - 2 - 7}}{{18}}} \right) + \left( {\dfrac{{21 + 10 + 4}}{{35}}} \right) + \dfrac{1}{{131}}\\
 =  - 1 + 1 + \dfrac{1}{{131}} = \dfrac{1}{{131}}\\
C4:\\
\dfrac{1}{3} - \dfrac{3}{4} + \dfrac{3}{5} + \dfrac{1}{{64}} - \dfrac{2}{9} - \dfrac{1}{{36}} + \dfrac{1}{{15}}\\
 = \left( {\dfrac{1}{3} + \dfrac{3}{5} + \dfrac{1}{{15}}} \right) + \left( { - \dfrac{3}{4} - \dfrac{2}{9} - \dfrac{1}{{36}}} \right) + \dfrac{1}{{64}}\\
 = \dfrac{{5 + 9 + 1}}{{15}} - \dfrac{{27 + 8 + 1}}{{36}} + \dfrac{1}{{64}}\\
 = 1 - 1 + \dfrac{1}{{64}}\\
 = \dfrac{1}{{64}}\\
C5:\\
\left( {\dfrac{1}{3} - \dfrac{1}{3}} \right) + \left( { - \dfrac{3}{5} + \dfrac{3}{5}} \right) + \left( {\dfrac{5}{7} - \dfrac{5}{7}} \right) + \left( { - \dfrac{7}{9} + \dfrac{7}{9}} \right) + \left( {\dfrac{9}{{11}} - \dfrac{9}{{11}}} \right) + \left( { - \dfrac{{11}}{{13}} + \dfrac{{11}}{{13}}} \right) + \dfrac{{13}}{{15}}\\
 = \dfrac{{13}}{{15}}
\end{array}\)