Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
e,\\
{\left( {{x^4}} \right)^3} = \dfrac{{{x^{18}}}}{{{x^7}}}\\
 \Leftrightarrow {x^{4.3}} = {x^{18 - 7}}\\
 \Leftrightarrow {x^{12}} = {x^{11}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
f,\\
x:\dfrac{3}{8}:\dfrac{5}{8} = x\\
 \Leftrightarrow x:\dfrac{3}{8}:\dfrac{5}{8} - x = 0\\
 \Leftrightarrow x\left( {1:\dfrac{3}{8}:\dfrac{5}{8} - 1} \right) = 0\\
 \Leftrightarrow x = 0\\
g,\\
{\left( {x - 1} \right)^5} =  - 32\\
 \Leftrightarrow {\left( {x - 1} \right)^5} = {\left( { - 2} \right)^5}\\
 \Leftrightarrow x - 1 =  - 2\\
 \Leftrightarrow x =  - 1\\
h,\,\,\,\,\,\,\,\,\,{5^{x + 1}} + {5^{x + 2}} = 750\\
 \Leftrightarrow {5^{x + 1}} + {5.5^{x + 1}} = 750\\
 \Leftrightarrow {6.5^{x + 1}} = 750\\
 \Leftrightarrow {5^{x + 1}} = 125\\
 \Leftrightarrow {5^{x + 1}} = {5^3}\\
 \Leftrightarrow x + 1 = 3\\
 \Leftrightarrow x = 2\\
i,\,\,\,\,\,\,\,\,\,\,\dfrac{3}{4} - \dfrac{2}{3}x = \dfrac{7}{5}\\
 \Leftrightarrow \dfrac{2}{3}x = \dfrac{3}{4} - \dfrac{7}{5}\\
 \Leftrightarrow \dfrac{2}{3}x =  - \dfrac{{13}}{{20}}\\
 \Leftrightarrow x = \left( { - \dfrac{{13}}{{20}}} \right):\dfrac{2}{3}\\
 \Leftrightarrow x =  - \dfrac{{39}}{{40}}\\
h,\,\,\,\,\,\,\,\,x + \dfrac{7}{6} = \dfrac{9}{5}x - \dfrac{4}{3}\\
 \Leftrightarrow \dfrac{7}{6} + \dfrac{4}{3} = \dfrac{9}{5}x - x\\
 \Leftrightarrow \dfrac{5}{2} = \dfrac{4}{5}x\\
 \Leftrightarrow x = \dfrac{5}{2}:\dfrac{4}{5}\\
 \Leftrightarrow x = \dfrac{{25}}{8}\\
l,\,\,\,\,\,\,\,\,\,\,\, - \dfrac{2}{3}\left| {x - 5} \right| =  - \dfrac{7}{6}\\
 \Leftrightarrow \left| {x - 5} \right| = \left( { - \dfrac{7}{6}} \right):\left( { - \dfrac{2}{3}} \right)\\
 \Leftrightarrow \left| {x - 5} \right| = \dfrac{7}{4}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 5 = \dfrac{7}{4}\\
x - 5 =  - \dfrac{7}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{4} + 5\\
x =  - \dfrac{7}{4} + 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{27}}{4}\\
x = \dfrac{{13}}{4}
\end{array} \right.\\
m,\,\,\,\,\,\,\,\,\,\dfrac{{81}}{{{3^{2x + 1}}}} = 3\\
 \Leftrightarrow {3^{2x + 1}} = 81:3\\
 \Leftrightarrow {3^{2x + 1}} = {3^3}\\
 \Leftrightarrow 2x + 1 = 3\\
 \Leftrightarrow x = 1\\
n,\,\,\,\,\,\,\dfrac{{x - 1}}{{x - 5}} = \dfrac{6}{7}\,\,\,\,\,\,\,\,\,\,\left( {x \ne 5} \right)\\
 \Leftrightarrow 7\left( {x - 1} \right) = 6\left( {x - 5} \right)\\
 \Leftrightarrow 7x - 7 = 6x - 30\\
 \Leftrightarrow x =  - 23\\
o,\,\,\,\,\,\,\,\,\,\dfrac{{x + 2}}{{ - 18}} = \dfrac{{ - 8}}{{x + 2}}\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = \left( { - 18} \right).\left( { - 8} \right)\\
 \Leftrightarrow {\left( {x + 2} \right)^2} = {12^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 2 = 12\\
x + 2 =  - 12
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 10\\
x =  - 14
\end{array} \right.
\end{array}\)