Đáp án:

\(\begin{array}{l}
1,\,\,\,\,x = \dfrac{1}{{81}}\\
2,\,\,\,\,x = \dfrac{{16}}{{25}}\\
3,\,\,\,\,\left[ \begin{array}{l}
x =  - \dfrac{1}{4}\\
x =  - \dfrac{3}{4}
\end{array} \right.\\
4,\,\,\,\,x =  - \dfrac{4}{3}\\
5,\,\,\,\,x = \dfrac{1}{8}\\
6,\,\,\,\,x =  - \dfrac{1}{3}\\
7,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
8,\,\,\,\,x = 1\\
9,\,\,\,\,x =  - 3,7\\
10,\,\,\,\,x = 1\\
11,\,\,\,\,\left[ \begin{array}{l}
x =  - \dfrac{1}{{15}}\\
x = \dfrac{1}{3}
\end{array} \right.\\
12,\,\,\,\,\left[ \begin{array}{l}
x = 4\\
x =  - 3
\end{array} \right.\\
13,\,\,\,\,x = 1\\
14,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{{13}}{5}\\
x = \dfrac{{ - 7}}{5}
\end{array} \right.\\
15,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{{11}}{{12}}\\
x =  - \dfrac{5}{{12}}
\end{array} \right.\\
16,\,\,\,\,Phương\,\,\,trình\,\,\,vô\,\,\,nghiệm
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
x:{\left( { - \dfrac{1}{3}} \right)^3} =  - \dfrac{1}{3}\\
 \Leftrightarrow x = \left( { - \dfrac{1}{3}} \right).{\left( { - \dfrac{1}{3}} \right)^3}\\
 \Leftrightarrow x = {\left( { - \dfrac{1}{3}} \right)^4}\\
 \Leftrightarrow x = {\left( {\dfrac{1}{3}} \right)^4}\\
 \Leftrightarrow x = \dfrac{1}{{81}}\\
2,\\
{\left( {\dfrac{4}{5}} \right)^5}.x = {\left( {\dfrac{4}{5}} \right)^7}\\
 \Leftrightarrow x = {\left( {\dfrac{4}{5}} \right)^7}:{\left( {\dfrac{4}{5}} \right)^5}\\
 \Leftrightarrow x = {\left( {\dfrac{4}{5}} \right)^{7 - 5}}\\
 \Leftrightarrow x = {\left( {\dfrac{4}{5}} \right)^2}\\
 \Leftrightarrow x = \dfrac{{16}}{{25}}\\
3,\\
{\left( {x + \dfrac{1}{2}} \right)^2} = \dfrac{1}{{16}}\\
 \Leftrightarrow {\left( {x + \dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{4}} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x + \dfrac{1}{2} = \dfrac{1}{4}\\
x + \dfrac{1}{2} =  - \dfrac{1}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4} - \dfrac{1}{2}\\
x =  - \dfrac{1}{4} - \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{4}\\
x =  - \dfrac{3}{4}
\end{array} \right.\\
4,\\
{\left( {3x + 1} \right)^3} =  - 27\\
 \Leftrightarrow {\left( {3x + 1} \right)^3} = {\left( { - 3} \right)^3}\\
 \Leftrightarrow 3x + 1 =  - 3\\
 \Leftrightarrow 3x =  - 3 - 1\\
 \Leftrightarrow 3x =  - 4\\
 \Leftrightarrow x =  - \dfrac{4}{3}\\
5,\\
{\left( {\dfrac{1}{2}} \right)^2}.x = {\left( {\dfrac{1}{2}} \right)^5}\\
 \Leftrightarrow x = {\left( {\dfrac{1}{2}} \right)^5}:{\left( {\dfrac{1}{2}} \right)^2}\\
 \Leftrightarrow x = {\left( {\dfrac{1}{2}} \right)^{5 - 2}}\\
 \Leftrightarrow x = {\left( {\dfrac{1}{2}} \right)^3}\\
 \Leftrightarrow x = \dfrac{1}{8}\\
6,\\
{\left( { - \dfrac{1}{3}} \right)^3}.x = \dfrac{1}{{81}}\\
 \Leftrightarrow {\left( { - \dfrac{1}{3}} \right)^3}.x = {\left( { - \dfrac{1}{3}} \right)^4}\\
 \Leftrightarrow x = {\left( { - \dfrac{1}{3}} \right)^4}:{\left( { - \dfrac{1}{3}} \right)^3}\\
 \Leftrightarrow x =  - \dfrac{1}{3}\\
7,\\
{\left( {2x - 3} \right)^2} = 16\\
 \Leftrightarrow {\left( {2x - 3} \right)^2} = {4^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 3 = 4\\
2x - 3 =  - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 4 + 3\\
2x =  - 4 + 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 7\\
2x =  - 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x =  - \dfrac{1}{2}
\end{array} \right.\\
8,\\
{\left( {x - \dfrac{2}{3}} \right)^3} = \dfrac{1}{{27}}\\
 \Leftrightarrow {\left( {x - \dfrac{2}{3}} \right)^3} = {\left( {\dfrac{1}{3}} \right)^3}\\
 \Leftrightarrow x - \dfrac{2}{3} = \dfrac{1}{3}\\
 \Leftrightarrow x = \dfrac{1}{3} + \dfrac{2}{3}\\
 \Leftrightarrow x = 1\\
9,\\
{\left( {x + 0,7} \right)^3} =  - 27\\
 \Leftrightarrow {\left( {x + 0,7} \right)^3} = {\left( { - 3} \right)^3}\\
 \Leftrightarrow x + 0,7 =  - 3\\
 \Leftrightarrow x =  - 3 - 0,7\\
 \Leftrightarrow x =  - 3,7\\
10,\\
{\left( {\dfrac{2}{3}x - \dfrac{1}{3}} \right)^5} = \dfrac{1}{{243}}\\
 \Leftrightarrow {\left( {\dfrac{2}{3}x - \dfrac{1}{3}} \right)^5} = \dfrac{1}{{{3^5}}}\\
 \Leftrightarrow {\left( {\dfrac{2}{3}x - \dfrac{1}{3}} \right)^5} = {\left( {\dfrac{1}{3}} \right)^5}\\
 \Leftrightarrow \dfrac{2}{3}x - \dfrac{1}{3} = \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{2}{3}x = \dfrac{1}{3} + \dfrac{1}{3}\\
 \Leftrightarrow \dfrac{2}{3}x = \dfrac{2}{3}\\
 \Leftrightarrow x = \dfrac{2}{3}:\dfrac{2}{3}\\
 \Leftrightarrow x = 1\\
11,\\
{\left( {\dfrac{2}{5} - 3x} \right)^2} = \dfrac{9}{{25}}\\
 \Leftrightarrow {\left( {\dfrac{2}{5} - 3x} \right)^2} = {\left( {\dfrac{3}{5}} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{2}{5} - 3x = \dfrac{3}{5}\\
\dfrac{2}{5} - 3x =  - \dfrac{3}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{2}{5} - \dfrac{3}{5}\\
3x = \dfrac{2}{5} - \left( { - \dfrac{3}{5}} \right)
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
3x =  - \dfrac{1}{5}\\
3x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{{15}}\\
x = \dfrac{1}{3}
\end{array} \right.\\
12,\\
{\left( {2x - 1} \right)^{10}} = {49^5}\\
 \Leftrightarrow {\left( {2x - 1} \right)^{10}} = {\left( {{7^2}} \right)^5}\\
 \Leftrightarrow {\left( {2x - 1} \right)^{10}} = {7^{2.5}}\\
 \Leftrightarrow {\left( {2x - 1} \right)^{10}} = {7^{10}}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 1 = 7\\
2x - 1 =  - 7
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 7 + 1\\
2x =  - 7 + 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 8\\
2x =  - 6
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 3
\end{array} \right.\\
13,\\
x:{5^2} = {\left( {\dfrac{3}{5}} \right)^2}:{3^2}\\
 \Leftrightarrow x:{5^2} = {\left( {\dfrac{3}{5}:3} \right)^2}\\
 \Leftrightarrow x:{5^2} = {\left( {\dfrac{1}{5}} \right)^2}\\
 \Leftrightarrow x = {\left( {\dfrac{1}{5}} \right)^2}{.5^2}\\
 \Leftrightarrow x = {\left( {\dfrac{1}{5}.5} \right)^2}\\
 \Leftrightarrow x = {1^2}\\
 \Leftrightarrow x = 1\\
14,\\
{\left( {x - \dfrac{3}{5}} \right)^2} = 4\\
 \Leftrightarrow {\left( {x - \dfrac{3}{5}} \right)^2} = {2^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{3}{5} = 2\\
x - \dfrac{3}{5} =  - 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 2 + \dfrac{3}{5}\\
x =  - 2 + \dfrac{3}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{13}}{5}\\
x = \dfrac{{ - 7}}{5}
\end{array} \right.\\
15,\\
{\left( {x - \dfrac{1}{4}} \right)^2} = \dfrac{4}{9}\\
 \Leftrightarrow {\left( {x - \dfrac{1}{4}} \right)^2} = {\left( {\dfrac{2}{3}} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{4} = \dfrac{2}{3}\\
x - \dfrac{1}{4} =  - \dfrac{2}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{2}{3} + \dfrac{1}{4}\\
x =  - \dfrac{2}{3} + \dfrac{1}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{{12}}\\
x =  - \dfrac{5}{{12}}
\end{array} \right.\\
16,\\
{\left( {2x - 5} \right)^4} =  - 81\\
{\left( {2x - 5} \right)^4} \ge 0 >  - 81,\,\,\,\forall x\\
 \Rightarrow Phương\,\,\,trình\,\,\,vô\,\,\,nghiệm
\end{array}\)