Đáp án:

$\begin{array}{l}
a)\frac{{x + 52}}{{21}} + \frac{{x + 51}}{{22}} = \frac{{x + 50}}{{23}} + \frac{{x + 49}}{{24}}\\
 \Rightarrow \frac{{x + 52}}{{21}} + 1 + \frac{{x + 51}}{{22}} + 1 = \frac{{x + 50}}{{23}} + 1 + \frac{{x + 49}}{{24}} + 1\\
 \Rightarrow \frac{{x + 52 + 21}}{{21}} + \frac{{x + 51 + 22}}{{22}} = \frac{{x + 50 + 23}}{{23}} + \frac{{x + 49 + 24}}{{24}}\\
 \Rightarrow \frac{{x + 73}}{{21}} + \frac{{x + 73}}{{22}} = \frac{{x + 73}}{{23}} + \frac{{x + 73}}{{24}}\\
 \Rightarrow \left( {x + 73} \right)\left( {\frac{1}{{21}} + \frac{1}{{22}} - \frac{1}{{23}} - \frac{1}{{24}}} \right) = 0\\
 \Rightarrow x + 73 = 0\\
 \Rightarrow x =  - 73\\
Vậy\,x =  - 73\\
b)A = \frac{3}{2} + \frac{3}{{{2^2}}} + \frac{3}{{{2^3}}} + ... + \frac{3}{{{2^{2020}}}}\\
A = 3.\left( {\frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}}} \right)\\
 \Rightarrow \frac{A}{3} = \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^{2020}}}}\\
 \Rightarrow 2.\frac{A}{3} = 1 + \frac{1}{2} + \frac{1}{{{2^2}}} + ... + \frac{1}{{{2^{2019}}}}\\
 \Rightarrow \frac{{2A}}{3} - \frac{A}{3} = 1 - \frac{1}{{{2^{2020}}}} = \frac{{{2^{2020}} - 1}}{{{2^{2020}}}}\\
 \Rightarrow \frac{A}{3} = \frac{{{2^{2020}} - 1}}{{{2^{2020}}}}\\
 \Rightarrow A = 3.\frac{{{2^{2020}} - 1}}{{{2^{2020}}}}
\end{array}$