Đáp án:

$\begin{array}{l}
1)x:{\left( { - \dfrac{1}{3}} \right)^3} =  - \dfrac{1}{3}\\
 \Rightarrow x:\left( { - \dfrac{1}{{27}}} \right) =  - \dfrac{1}{3}\\
 \Rightarrow x =  - \dfrac{1}{3}.\left( { - \dfrac{1}{{27}}} \right)\\
 \Rightarrow x = \dfrac{1}{{81}}\\
\text{Vậy}\,x = \dfrac{1}{{81}}\\
2){\left( {\dfrac{4}{5}} \right)^5}.x = {\left( {\dfrac{4}{5}} \right)^7}\\
 \Rightarrow x = {\left( {\dfrac{4}{5}} \right)^7}:{\left( {\dfrac{4}{5}} \right)^5}\\
 \Rightarrow x = {\left( {\dfrac{4}{5}} \right)^2}\\
 \Rightarrow x = \dfrac{{16}}{{25}}\\
\text{Vậy}\,x = \dfrac{{16}}{{25}}\\
3)\\
{\left( {x + \dfrac{1}{2}} \right)^2} = \dfrac{1}{{16}}\\
 \Rightarrow \left[ \begin{array}{l}
x + \dfrac{1}{2} = \dfrac{1}{4}\\
x + \dfrac{1}{2} =  - \dfrac{1}{4}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4} - \dfrac{1}{2}\\
x =  - \dfrac{1}{4} - \dfrac{1}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 1}}{4}\\
x =  - \dfrac{3}{4}
\end{array} \right.\\
\text{Vậy}\,x =  - \dfrac{1}{4};x = \dfrac{{ - 3}}{4}\\
4){\left( {3x + 1} \right)^3} =  - 27\\
 \Rightarrow 3x + 1 =  - 3\\
 \Rightarrow 3x =  - 4\\
 \Rightarrow x =  - \dfrac{4}{3}\\
\text{Vậy}\,x =  - \dfrac{4}{3}\\
5){\left( {\dfrac{1}{2}} \right)^2}.x = {\left( {\dfrac{1}{2}} \right)^5}\\
 \Rightarrow x = {\left( {\dfrac{1}{2}} \right)^5}:{\left( {\dfrac{1}{2}} \right)^2}\\
 \Rightarrow x = {\left( {\dfrac{1}{2}} \right)^3}\\
 \Rightarrow x = \dfrac{1}{8}\\
\text{Vậy}\,x = \dfrac{1}{8}\\
6){\left( { - \dfrac{1}{3}} \right)^3}.x = \dfrac{1}{{81}}\\
 \Rightarrow {\left( { - \dfrac{1}{3}} \right)^3}.x = {\left( { - \dfrac{1}{3}} \right)^4}\\
 \Rightarrow x = {\left( { - \dfrac{1}{3}} \right)^4}:{\left( { - \dfrac{1}{3}} \right)^3}\\
 \Rightarrow x =  - \dfrac{1}{3}\\
\text{Vậy}\,x =  - \dfrac{1}{3}\\
9){\left( {x + 0,7} \right)^3} =  - 27\\
 \Rightarrow x + 0,7 =  - 3\\
 \Rightarrow x =  - 3 - 0,7\\
 \Rightarrow x =  - 3,7\\
\text{Vậy}\,x =  - 3,7
\end{array}$

$\begin{array}{l}
10){\left( {\dfrac{2}{3}x - \dfrac{1}{3}} \right)^5} = \dfrac{1}{{243}}\\
 \Rightarrow {\left( {\dfrac{2}{3}x - \dfrac{1}{3}} \right)^5} = {\left( {\dfrac{1}{3}} \right)^5}\\
 \Rightarrow \dfrac{2}{3}x - \dfrac{1}{3} = \dfrac{1}{3}\\
 \Rightarrow \dfrac{2}{3}x = \dfrac{2}{3}\\
 \Rightarrow x = 1\\
\text{Vậy}\,x = 1\\
11){\left( {\dfrac{2}{5} - 3x} \right)^2} = \dfrac{9}{{25}}\\
 \Rightarrow \left[ \begin{array}{l}
\dfrac{2}{5} - 3x = \dfrac{3}{5}\\
\dfrac{2}{5} - 3x =  - \dfrac{3}{5}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
3x =  - \dfrac{1}{5}\\
3x = 1
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{{15}}\\
x = \dfrac{1}{3}
\end{array} \right.\\
\text{Vậy}\,x =  - \dfrac{1}{{15}};x = \dfrac{1}{3}\\
12)\\
{\left( {2x - 1} \right)^{10}} = {49^5}\\
 \Rightarrow {\left( {2x - 1} \right)^{10}} = {\left( {{7^2}} \right)^5}\\
 \Rightarrow {\left( {2x - 1} \right)^{10}} = {7^{10}}\\
 \Rightarrow \left[ \begin{array}{l}
2x - 1 = 7\\
2x - 1 =  - 7
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2x = 8\\
2x =  - 6
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = 4\\
x =  - 3
\end{array} \right.\\
13)\\
x:{5^2} = {\left( {\dfrac{3}{5}} \right)^2}:{3^2}\\
 \Rightarrow x:25 = {\left( {\dfrac{3}{5}:3} \right)^2}\\
 \Rightarrow x:25 = {\left( {\dfrac{1}{5}} \right)^2}\\
 \Rightarrow x = \dfrac{1}{{25}}.25\\
 \Rightarrow x = 1\\
\text{Vậy}\,x = 1\\
14){\left( {x - \dfrac{3}{5}} \right)^2} = 4\\
 \Rightarrow \left[ \begin{array}{l}
x - \dfrac{3}{5} = 2\\
x - \dfrac{3}{5} =  - 2
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{13}}{5}\\
x = \dfrac{{ - 7}}{5}
\end{array} \right.\\
\text{Vậy}\,x = \dfrac{{13}}{5};x =  - \dfrac{7}{5}
\end{array}$