Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
b,\\
{\left( {x - 1} \right)^3} - {3^5}:{3^4} + {2.2^3} = 14\\
 \Leftrightarrow {\left( {x - 1} \right)^3} - {3^{5 - 4}} + 2.8 = 14\\
 \Leftrightarrow {\left( {x - 1} \right)^3} - {3^1} + 16 = 14\\
 \Leftrightarrow {\left( {x - 1} \right)^3} - 3 + 16 = 14\\
 \Leftrightarrow {\left( {x - 1} \right)^3} = 14 + 3 - 16\\
 \Leftrightarrow {\left( {x - 1} \right)^3} = 1\\
 \Leftrightarrow x - 1 = 1\\
 \Leftrightarrow x = 1 + 1\\
 \Leftrightarrow x = 2\\
c,\\
A = 1 + 2 + {2^2} + {2^3} + ..... + {2^{2020}}\\
 \Leftrightarrow 2A = 2.\left( {1 + 2 + {2^2} + {2^3} + ..... + {2^{2020}}} \right)\\
 \Leftrightarrow 2A = 2.1 + 2.2 + {2.2^2} + {2.2^3} + .... + {2.2^{2020}}\\
 \Leftrightarrow 2A = 2 + {2^2} + {2^3} + {2^4} + ..... + {2^{2021}}\\
 \Leftrightarrow 2A - A = \left( {2 + {2^2} + {2^3} + {2^4} + ..... + {2^{2021}}} \right) - \left( {1 + 2 + {2^2} + {2^3} + ..... + {2^{2020}}} \right)\\
 \Leftrightarrow A = {2^{2021}} - 1\\
A = {2^{x - 1}} - 1\\
 \Leftrightarrow {2^{2021}} - 1 = {2^{x - 1}} - 1\\
 \Leftrightarrow {2^{2021}} = {2^{x - 1}}\\
 \Leftrightarrow x - 1 = 2021\\
 \Leftrightarrow x = 2021 + 1\\
 \Leftrightarrow x = 2022\\
d,\\
{5^{2x - 3}} - {2.5^2} = {5^2}.3\\
 \Leftrightarrow {5^{2x - 3}} = {5^2}.3 + {2.5^2}\\
 \Leftrightarrow {5^{2x - 3}} = {5^2}.\left( {3 + 2} \right)\\
 \Leftrightarrow {5^{2x - 3}} = {5^2}.5\\
 \Leftrightarrow {5^{2x - 3}} = {5^3}\\
 \Leftrightarrow 2x - 3 = 3\\
 \Leftrightarrow 2x = 3 + 3\\
 \Leftrightarrow 2x = 6\\
 \Leftrightarrow x = 6:2\\
 \Leftrightarrow x = 3
\end{array}\)