Đáp án:

\(\begin{array}{l}
1,\,\,\,\,x =  - \dfrac{{27}}{{125}}\\
2,\,\,\,\,x =  - \dfrac{1}{3}\\
3,\,\,\,\,x = 3\\
4,\,\,\,\,x = 3\\
5,\,\,\,\,x = 6\\
6,\,\,\,\,x = 2\\
7,\,\,\,\,x = 1\\
8,\,\,\,\,x = \dfrac{5}{2}\\
9,\,\,\,\,x =  - 2\\
10,\,\,\,\,x = 1\\
11,\,\,\,\,x =  \pm \dfrac{2}{5}\\
12,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x =  - \dfrac{7}{2}
\end{array} \right.\\
13,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{{27}}{8}\\
x =  - \dfrac{{27}}{8}
\end{array} \right.\\
14,\,\,\,\,x =  - 2\\
15,\,\,\,\,\left[ \begin{array}{l}
x = 4\\
x = 1
\end{array} \right.\\
16,\,\,\,\,x =  - 9\\
17,\,\,\,\,\left[ \begin{array}{l}
x = \dfrac{7}{9}\\
x = \dfrac{7}{9}
\end{array} \right.
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
x:{\left( { - \dfrac{3}{5}} \right)^2} =  - \dfrac{3}{5}\\
 \Leftrightarrow x = \left( { - \dfrac{3}{5}} \right).{\left( { - \dfrac{3}{5}} \right)^2}\\
 \Leftrightarrow x = {\left( { - \dfrac{3}{5}} \right)^3}\\
 \Leftrightarrow x =  - \dfrac{{{3^3}}}{{{5^3}}}\\
 \Leftrightarrow x =  - \dfrac{{27}}{{125}}\\
2,\\
{\left( { - \dfrac{1}{3}} \right)^3}.x = \dfrac{1}{{81}}\\
 \Leftrightarrow {\left( { - \dfrac{1}{3}} \right)^3}.x = \dfrac{1}{{{{\left( { - 3} \right)}^4}}}\\
 \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}\\
3,\\
{5^x} = 125\\
 \Leftrightarrow {5^x} = {5^3}\\
 \Leftrightarrow x = 3\\
4,\\
{\left( { - 7} \right)^x} =  - 343\\
 \Leftrightarrow {\left( { - 7} \right)^x} =  - {7^3}\\
 \Leftrightarrow {\left( { - 7} \right)^x} = {\left( { - 7} \right)^3}\\
 \Leftrightarrow x = 3\\
5,\\
{\left( { - \dfrac{1}{7}} \right)^x} = {\left( { - \dfrac{1}{{343}}} \right)^2}\\
 \Leftrightarrow {\left( { - \dfrac{1}{7}} \right)^x} = {\left( { - \dfrac{1}{{{7^3}}}} \right)^2}\\
 \Leftrightarrow {\left( { - \dfrac{1}{7}} \right)^x} = {\left[ {{{\left( { - \dfrac{1}{7}} \right)}^3}} \right]^2}\\
 \Leftrightarrow {\left( { - \dfrac{1}{7}} \right)^x} = {\left( { - \dfrac{1}{7}} \right)^6}\\
 \Leftrightarrow x = 6\\
6,\\
{\left( { - \dfrac{5}{2}} \right)^x} = \dfrac{{25}}{4}\\
 \Leftrightarrow {\left( { - \dfrac{5}{2}} \right)^x} = \dfrac{{{{\left( { - 5} \right)}^2}}}{{{2^2}}}\\
 \Leftrightarrow {\left( { - \dfrac{5}{2}} \right)^x} = {\left( { - \dfrac{5}{2}} \right)^2}\\
 \Leftrightarrow x = 2\\
7,\\
{\left( {\dfrac{2}{7}} \right)^{3x}} = \dfrac{8}{{343}}\\
 \Leftrightarrow {\left( {\dfrac{2}{7}} \right)^{3x}} = \dfrac{{{2^3}}}{{{7^3}}}\\
 \Leftrightarrow {\left( {\dfrac{2}{7}} \right)^{3x}} = {\left( {\dfrac{2}{7}} \right)^3}\\
 \Leftrightarrow 3x = 3\\
 \Leftrightarrow x = 1\\
8,\\
{\left( {\dfrac{1}{2}} \right)^{2x - 1}} = \dfrac{1}{{16}}\\
 \Leftrightarrow {\left( {\dfrac{1}{2}} \right)^{2x - 1}} = \dfrac{1}{{{2^4}}}\\
 \Leftrightarrow {\left( {\dfrac{1}{2}} \right)^{2x - 1}} = {\left( {\dfrac{1}{2}} \right)^4}\\
 \Leftrightarrow 2x - 1 = 4\\
 \Leftrightarrow 2x = 4 + 1\\
 \Leftrightarrow 2x = 5\\
 \Leftrightarrow x = \dfrac{5}{2}\\
9,\\
{5^{x - 1}} = \dfrac{1}{{125}}\\
 \Leftrightarrow {5^{x - 1}}.125 = 1\\
 \Leftrightarrow {5^{x - 1}}{.5^3} = 1\\
 \Leftrightarrow {5^{\left( {x - 1} \right) + 3}} = 1\\
 \Leftrightarrow {5^{x + 2}} = {5^0}\\
 \Leftrightarrow x + 2 = 0\\
 \Leftrightarrow x =  - 2\\
10,\\
\dfrac{1}{9}{.27^x} = {3^x}\\
 \Leftrightarrow {27^x} = {9.3^x}\\
 \Leftrightarrow {\left( {{3^3}} \right)^x} = {3^2}{.3^x}\\
 \Leftrightarrow {3^{3x}} = {3^{x + 2}}\\
 \Leftrightarrow 3x = x + 2\\
 \Leftrightarrow 2x = 2\\
 \Leftrightarrow x = 1\\
11,\\
{x^4} = \dfrac{{16}}{{625}}\\
 \Leftrightarrow {x^4} = \dfrac{{{2^4}}}{{{5^4}}}\\
 \Leftrightarrow {x^4} = {\left( {\dfrac{2}{5}} \right)^4}\\
 \Leftrightarrow x =  \pm \dfrac{2}{5}\\
12,\\
{\left( {2x.\dfrac{1}{{13}}} \right)^2} = \dfrac{{49}}{{169}}\\
 \Leftrightarrow {\left( {\dfrac{{2x}}{{13}}} \right)^2} = \dfrac{{{7^2}}}{{{{13}^2}}}\\
 \Leftrightarrow {\left( {\dfrac{{2x}}{{13}}} \right)^2} = {\left( {\dfrac{7}{{13}}} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{{2x}}{{13}} = \dfrac{7}{{13}}\\
\dfrac{{2x}}{{13}} =  - \dfrac{7}{{13}}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 7\\
2x =  - 7
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{2}\\
x =  - \dfrac{7}{2}
\end{array} \right.\\
13,\\
{\left( {\dfrac{2}{5}.x} \right)^2} = \dfrac{{81}}{{16}}\\
 \Leftrightarrow {\left( {\dfrac{2}{5}x} \right)^2} = {\left( {\dfrac{9}{4}} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
\dfrac{2}{5}x = \dfrac{9}{4}\\
\dfrac{2}{5}x =  - \dfrac{9}{4}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{9}{4}:\dfrac{2}{5}\\
x = \left( { - \dfrac{9}{4}} \right):\dfrac{2}{5}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{27}}{8}\\
x =  - \dfrac{{27}}{8}
\end{array} \right.\\
14,\\
{\left( {x - 1} \right)^3} =  - 27\\
 \Leftrightarrow {\left( {x - 1} \right)^3} = {\left( { - 3} \right)^3}\\
 \Leftrightarrow x - 1 =  - 3\\
 \Leftrightarrow x =  - 3 + 1\\
 \Leftrightarrow x =  - 2\\
15,\\
{\left( {2x - 5} \right)^4} = 81\\
 \Leftrightarrow {\left( {2x - 5} \right)^4} = {3^4}\\
 \Leftrightarrow \left[ \begin{array}{l}
2x - 5 = 3\\
2x - 5 =  - 3
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 3 + 5\\
2x =  - 3 + 5
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
2x = 8\\
2x = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 1
\end{array} \right.\\
16,\\
{\left( {x + 5} \right)^3} =  - 64\\
 \Leftrightarrow {\left( {x + 5} \right)^3} =  - {4^3}\\
 \Leftrightarrow {\left( {x + 5} \right)^3} = {\left( { - 4} \right)^3}\\
 \Leftrightarrow x + 5 =  - 4\\
 \Leftrightarrow x =  - 4 - 5\\
 \Leftrightarrow x =  - 9\\
17,\\
{\left( {3x - 2} \right)^2} = \dfrac{1}{9}\\
 \Leftrightarrow {\left( {3x - 2} \right)^2} = {\left( {\dfrac{1}{3}} \right)^2}\\
 \Leftrightarrow \left[ \begin{array}{l}
3x - 2 = \dfrac{1}{3}\\
3x - 2 =  - \dfrac{1}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{1}{3} + 2\\
3x =  - \dfrac{1}{3} + 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
3x = \dfrac{7}{3}\\
3x = \dfrac{5}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{7}{9}\\
x = \dfrac{7}{9}
\end{array} \right.
\end{array}\)