Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
4,\\
a,\\
{\left( {2x + 1} \right)^3} = 9.81\\
 \Leftrightarrow {\left( {2x + 1} \right)^3} = {9.9^2}\\
 \Leftrightarrow {\left( {2x + 1} \right)^3} = {9^{1 + 2}}\\
 \Leftrightarrow {\left( {2x + 1} \right)^3} = {9^3}\\
 \Leftrightarrow 2x + 1 = 9\\
 \Leftrightarrow 2x = 9 - 1\\
 \Leftrightarrow 2x = 8\\
 \Leftrightarrow x = 8:2\\
 \Leftrightarrow x = 4\\
c,\\
{\left( {x - 5} \right)^7} = {\left( {x - 5} \right)^3}\\
 \Leftrightarrow {\left( {x - 5} \right)^7} - {\left( {x - 5} \right)^3} = 0\\
 \Leftrightarrow {\left( {x - 5} \right)^{3 + 4}} - {\left( {x - 5} \right)^3} = 0\\
 \Leftrightarrow {\left( {x - 5} \right)^3}.{\left( {x - 5} \right)^4} - {\left( {x - 5} \right)^3} = 0\\
 \Leftrightarrow {\left( {x - 5} \right)^3}.\left[ {{{\left( {x - 5} \right)}^4} - 1} \right] = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{\left( {x - 5} \right)^3} = 0\\
{\left( {x - 5} \right)^4} = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 5 = 0\\
x - 5 = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = 6
\end{array} \right.\\
d,\\
{2^x}.4 = 128\\
 \Leftrightarrow {2^x} = 128:4\\
 \Leftrightarrow {2^x} = 32\\
 \Leftrightarrow {2^x} = {2^5}\\
 \Leftrightarrow x = 5\\
e,\\
3.{\left( {3x - 4} \right)^3} = 375\\
 \Leftrightarrow {\left( {3x - 4} \right)^3} = 375:3\\
 \Leftrightarrow {\left( {3x - 4} \right)^3} = 125\\
 \Leftrightarrow {\left( {3x - 4} \right)^3} = {5^3}\\
 \Leftrightarrow 3x - 4 = 5\\
 \Leftrightarrow 3x = 5 + 4\\
 \Leftrightarrow 3x = 9\\
 \Leftrightarrow x = 9:3\\
 \Leftrightarrow x = 3\\
5,\\
a,\\
{8.2^5}.16 = {2^3}{.2^5}{.2^4} = {2^{3 + 5 + 4}} = {2^{12}}\\
b,\\
{27^{16}}{.9^{10}} = {\left( {{3^3}} \right)^{16}}.{\left( {{3^2}} \right)^{10}} = {3^{3.16}}{.3^{2.10}} = {3^{48}}{.3^{20}} = {3^{48 + 20}} = {3^{68}}\\
c,\\
{625^5}:{25^7} = {\left( {{5^4}} \right)^5}:{\left( {{5^2}} \right)^7} = {5^{4.5}}:{5^{2.7}} = {5^{20}}:{5^{14}} = {5^{20 - 14}} = {5^6}\\
d,\\
\dfrac{{{3^{10}}.11 + {3^8}.45}}{{{3^9}.16}} = \dfrac{{{3^{10}}.11 + {3^8}{{.3}^2}.5}}{{{3^9}.16}} = \dfrac{{{3^{10}}.11 + {3^{8 + 2}}.5}}{{{3^9}.16}}\\
 = \dfrac{{{3^{10}}.11 + {3^{10}}.5}}{{{3^9}.16}} = \dfrac{{{3^{10}}.\left( {11 + 5} \right)}}{{{3^9}.16}} = \dfrac{{{3^{10}}.16}}{{{3^9}.16}} = 3
\end{array}\)