Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
{\left( {3x - 1} \right)^2} = 25\\
 \Leftrightarrow {\left( {3x - 1} \right)^2} = {5^2}\\
 \Leftrightarrow 3x - 1 = 5\\
 \Leftrightarrow 3x = 5 + 1\\
 \Leftrightarrow 3x = 6\\
 \Leftrightarrow x = 6:3\\
 \Leftrightarrow x = 2\\
b,\\
{\left( {2x - 5} \right)^3} = 27\\
 \Leftrightarrow {\left( {2x - 5} \right)^3} = {3^3}\\
 \Leftrightarrow 2x - 5 = 3\\
 \Leftrightarrow 2x = 5 + 3\\
 \Leftrightarrow 2x = 8\\
 \Leftrightarrow x = 8:2\\
 \Leftrightarrow x = 4\\
c,\\
{3^x} = {9^8}{.27^9}{.81^2}\\
 \Leftrightarrow {3^x} = {\left( {{3^2}} \right)^8}.{\left( {{3^3}} \right)^9}.{\left( {{3^4}} \right)^2}\\
 \Leftrightarrow {3^x} = {3^{16}}{.3^{27}}{.3^8}\\
 \Leftrightarrow {3^x} = {3^{16 + 27 + 8}}\\
 \Leftrightarrow {3^x} = {3^{51}}\\
 \Leftrightarrow x = 51\\
d,\\
{3^{x + 5}}:{9^3} = {27^4}\\
 \Leftrightarrow {3^{x + 5}} = {9^3}{.27^4}\\
 \Leftrightarrow {3^{x + 5}} = {\left( {{3^2}} \right)^3}.{\left( {{3^3}} \right)^4}\\
 \Leftrightarrow {3^{x + 5}} = {3^6}{.3^{12}}\\
 \Leftrightarrow {3^{x + 5}} = {3^{6 + 12}}\\
 \Leftrightarrow {3^{x + 5}} = {3^{18}}\\
 \Leftrightarrow x + 5 = 18\\
 \Leftrightarrow x = 18 - 5\\
 \Leftrightarrow x = 13\\
e,\\
{2^{x + 2}} - {2^x} = 96\\
 \Leftrightarrow {2^x}{.2^2} - {2^x} = 96\\
 \Leftrightarrow {2^x}.\left( {{2^2} - 1} \right) = 96\\
 \Leftrightarrow {2^x}.3 = 96\\
 \Leftrightarrow {2^x} = 96:3\\
 \Leftrightarrow {2^x} = 32\\
 \Leftrightarrow {2^x} = {2^5}\\
 \Leftrightarrow x = 5\\
f,\\
{3^{x + 2}} + {4.3^{x + 1}} = {7.3^6}\\
 \Leftrightarrow {3^{x + 1}}.3 + {4.3^{x + 1}} = {7.3^6}\\
 \Leftrightarrow {7.3^{x + 1}} = {7.3^6}\\
 \Leftrightarrow x + 1 = 6\\
 \Leftrightarrow x = 5\\
g,\\
8 \le {8^{3x - 8}} \le {8^7}\\
 \Leftrightarrow {8^1} \le {8^{3x - 8}} \le {8^7}\\
 \Leftrightarrow 1 \le 3x - 8 \le 7\\
 \Leftrightarrow 1 + 8 \le 3x \le 7 + 8\\
 \Leftrightarrow 9 \le 3x \le 15\\
 \Leftrightarrow 3 \le x \le 5\\
 \Rightarrow x \in \left\{ {3;4;5} \right\}\\
h,\\
{\left( {2x - 3} \right)^{2020}} = {\left( {2x - 3} \right)^{2022}}\\
 \Leftrightarrow {\left( {2x - 3} \right)^{2020}} - {\left( {2x - 3} \right)^{2022}} = 0\\
 \Leftrightarrow {\left( {2x - 3} \right)^{2020}}.\left[ {1 - {{\left( {2x - 3} \right)}^2}} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\left( {2x - 3} \right)^{2020}} = 0\\
1 - {\left( {2x - 3} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 0\\
2x - 3 = 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
2x = 3\\
2x = 4
\end{array} \right. \Leftrightarrow x = 2\,\,\,\left( {x \in N} \right)
\end{array}\)

\(\begin{array}{l}
i,\\
{\left( {5 - x} \right)^{2021}} = {\left( {5 - x} \right)^{2018}}\\
 \Leftrightarrow {\left( {5 - x} \right)^{2021}} - {\left( {5 - x} \right)^{2018}} = 0\\
 \Leftrightarrow {\left( {5 - x} \right)^{2018}}.\left[ {{{\left( {5 - x} \right)}^3} - 1} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\left( {5 - x} \right)^{2018}} = 0\\
{\left( {5 - x} \right)^3} - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
5 - x = 0\\
5 - x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 4
\end{array} \right.
\end{array}\)