Đáp án:

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

Giải thích các bước giải:

 $1)
x:\left ( -\dfrac{1}{3} \right )^3=\dfrac{-1}{3}\\
\Leftrightarrow x=\dfrac{-1}{3}.\left (\dfrac{-1}{3}  \right )^3\\
\Leftrightarrow x=\left (\dfrac{-1}{3}  \right )^4=\dfrac{1}{81}\\
2)
\left ( \dfrac{4}{5} \right )^5x=\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 )^2\\
3)
\left ( x+\dfrac{1}{2} \right )^2=\dfrac{1}{16}\\
\Leftrightarrow x+\dfrac{1}{2} =\pm \dfrac{1}{4}\\
\Leftrightarrow {\left[\begin{aligned}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{4}-\dfrac{1}{2}\\x=-\dfrac{1}{4}-\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{1}{4}-\dfrac{2}{4}\\x=-\dfrac{1}{4}-\dfrac{2}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{-1}{4}\\x=-\dfrac{3}{4}\end{aligned}\right.}\\
4)
(3x+1)^3=-27\\
\Leftrightarrow (3x+1)^3=(-3)^3\\
\Leftrightarrow 3x+1=-3\\
\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 )^3=\dfrac{1}{8}\\
6)
\left ( -\dfrac{1}{3} \right )^3.x=\dfrac{1}{81}\\
\Leftrightarrow x=\dfrac{1}{81}.(-3)^3=\dfrac{-1}{3}\\
7)
(2x-3)^2=16\\
\Leftrightarrow 2x-3=\pm 4\\
\Leftrightarrow {\left[\begin{aligned}2x-3=4\\2x-3=-4\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=7\\2x=-1\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{7}{2}\\x=\dfrac{-1}{2}\end{aligned}\right.}\\
8)
\left ( x-\dfrac{2}{3} \right )^3=\dfrac{1}{27}\\
\Leftrightarrow \left ( x-\dfrac{2}{3} \right )^3=(\dfrac{1}{3})^3\\
\Leftrightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\\
\Leftrightarrow x=\dfrac{1}{3}+\dfrac{2}{3}=1\\
9)
(x+0,7)^3=-27\\
\Leftrightarrow (x+0,7)^3=(-3)^3\\
\Leftrightarrow x+0,7=-3\\
\Leftrightarrow x=-3-0,7=-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=\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}=\dfrac{2}{3}\\
\Leftrightarrow x=\dfrac{2}{3}.\dfrac{3}{2}=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 \dfrac{2}{5}-3x =\pm \dfrac{3}{5}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{2}{5}-3x =\dfrac{3}{5}\\\dfrac{2}{5}-3x =- \dfrac{3}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}3x=\dfrac{2}{5}-\dfrac{3}{5}\\3x=\dfrac{2}{5}+ \dfrac{3}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}3x=-\dfrac{1}{5}\\3x=\dfrac{5}{5}=1\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=-\dfrac{1}{15}\\x=\dfrac{1}{3}\end{aligned}\right.}\\
12)
(2x-1)^{10}=49^5\\
\Leftrightarrow (2x-1)^{10}=(7^2)^5=7^{10}\\
\Leftrightarrow 2x-1=\pm 7\\
\Leftrightarrow {\left[\begin{aligned}2x-1=7\\2x-1=-7\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}2x=8\\2x=-6\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=4\\x=-3\end{aligned}\right.}\\
14)
\left ( x-\dfrac{3}{5} \right )^2=4\\
\Leftrightarrow x-\dfrac{3}{5}=\pm 2\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{3}{5}=2\\x-\dfrac{3}{5}=- 2\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=2+\dfrac{3}{5}\\x=- 2+\dfrac{3}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{10}{5}+\dfrac{3}{5}\\x=\dfrac{-10}{5}+\dfrac{3}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{13}{5}\\x=\dfrac{-7}{5}\end{aligned}\right.}\\
13)
x:5^2=\left ( \dfrac{3}{5} \right )^2:3^2\\
\Leftrightarrow x:5^2=\dfrac{3^2}{5^2} .\dfrac{1}{3^2}=\dfrac{1}{5^2}\\
\Leftrightarrow x=\dfrac{1}{5^2}.5^2=1\\
15)
\left ( x-\dfrac{1}{4} \right )^2=\dfrac{4}{9}\\
\Leftrightarrow x-\dfrac{1}{4}=\pm \dfrac{2}{3}\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{1}{4}= \dfrac{2}{3}\\x-\dfrac{1}{4}=- \dfrac{2}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x= \dfrac{2}{3}+\dfrac{1}{4}\\x=- \dfrac{2}{3}+\dfrac{1}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x= \dfrac{8}{12}+\dfrac{3}{12}\\x=- \dfrac{8}{12}+\dfrac{3}{12}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x= \dfrac{11}{12}\\x=- \dfrac{5}{12}\end{aligned}\right.}\\
16)$
Phương trình vô nghiệm