Đáp án:
$\begin{array}{l}
B = \dfrac{1}{{1.4}} + \dfrac{1}{{4.7}} + \dfrac{1}{{7.10}} + ... + \dfrac{1}{{97.100}}\\
\Leftrightarrow 3B = \dfrac{3}{{1.4}} + \dfrac{3}{{4.7}} + \dfrac{3}{{7.10}} + ... + \dfrac{3}{{97.100}}\\
\Leftrightarrow 3B = 1 - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{7} + \dfrac{1}{7} - \dfrac{1}{{10}} + ... + \dfrac{1}{{97}} - \dfrac{1}{{100}}\\
\Leftrightarrow 3B = 1 - \dfrac{1}{{100}} = \dfrac{{99}}{{100}}\\
\Leftrightarrow B = \dfrac{{33}}{{100}}\\
C = \left( {1 - \dfrac{1}{2}} \right)\left( {1 - \dfrac{1}{3}} \right)\left( {1 - \dfrac{1}{4}} \right)...\left( {1 - \dfrac{1}{{100}}} \right)\\
= \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}....\dfrac{{99}}{{100}}\\
= \dfrac{1}{{100}}\\
2)a)\left| {3x - 1} \right| - \dfrac{1}{2} = \dfrac{{13}}{2}\\
\Leftrightarrow \left| {3x - 1} \right| = \dfrac{{13}}{2} + \dfrac{1}{2} = 7\\
\Leftrightarrow \left[ \begin{array}{l}
3x - 1 = 7\\
3x - 1 = - 7
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{8}{3}\\
x = - 2
\end{array} \right.\\
Vậy\,x = \dfrac{8}{3};x = - 2\\
b){\left( {2x - \dfrac{1}{3}} \right)^2} = \dfrac{4}{9}\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{3} = \dfrac{2}{3} \Leftrightarrow x = \dfrac{1}{2}\\
2x - \dfrac{1}{3} = - \dfrac{2}{3} \Leftrightarrow x = - \dfrac{1}{6}
\end{array} \right.\\
Vậy\,x = - \dfrac{1}{6};x = \dfrac{1}{2}\\
c)\dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{x\left( {x + 1} \right)}} = \dfrac{{2021}}{{2022}}\\
\Leftrightarrow 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + .... + \dfrac{1}{x} - \dfrac{1}{{x + 1}} = \dfrac{{2021}}{{2022}}\\
\Leftrightarrow 1 - \dfrac{1}{{x + 1}} = 1 - \dfrac{1}{{2022}}\\
\Leftrightarrow \dfrac{1}{{x + 1}} = \dfrac{1}{{2022}}\\
\Leftrightarrow x + 1 = 2022\\
\Leftrightarrow x = 2021\\
Vậy\,x = 2021
\end{array}$
$\begin{array}{l}
3)\dfrac{a}{b} = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{9} + \dfrac{1}{{10}}\\
= \left( {1 + \dfrac{1}{{10}}} \right) + \left( {\dfrac{1}{2} + \dfrac{1}{9}} \right) + \left( {\dfrac{1}{3} + \dfrac{1}{8}} \right) + \left( {\dfrac{1}{4} + \dfrac{1}{7}} \right)\\
+ \left( {\dfrac{1}{5} + \dfrac{1}{6}} \right)\\
= \dfrac{{11}}{{10}} + \dfrac{{11}}{{18}} + \dfrac{{11}}{{24}} + \dfrac{{11}}{{28}} + \dfrac{{11}}{{30}}\\
= 11.\left( {\dfrac{1}{{10}} + \dfrac{1}{{18}} + \dfrac{1}{{24}} + \dfrac{1}{{28}} + \dfrac{1}{{30}}} \right)\\
= 11.\dfrac{c}{b}\\
= \dfrac{{11.c}}{b} = \dfrac{a}{b}\\
\Leftrightarrow a = 11.c\\
\Leftrightarrow a \vdots 11
\end{array}$