Đáp án:

\(\begin{array}{l}
2,\\
A =  - \dfrac{1}{3}\\
3,\\
a,\,\,\,x =  \pm \dfrac{3}{5}\\
b,\,\,\,x = \dfrac{9}{5}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
{16^5} + {2^{15}} = {\left( {{2^4}} \right)^5} + {2^{15}} = {2^{4.5}} + {2^{15}} = {2^{20}} + {2^{15}}\\
 = {2^{15}}{.2^5} + {2^{15}} = {2^{15}}.\left( {{2^5} + 1} \right) = {2^{15}}.33\\
\left( {{2^{15}}.33} \right) \vdots 33\\
 \Rightarrow \left( {{{16}^5} + {2^{15}}} \right) \vdots 33\\
2,\\
A = \dfrac{{{4^5}{{.9}^4} - {{2.6}^9}}}{{{2^{10}}{{.3}^8} + {6^8}.20}}\\
 = \dfrac{{{{\left( {{2^2}} \right)}^5}.{{\left( {{3^2}} \right)}^4} - 2.{{\left( {2.3} \right)}^9}}}{{{2^{10}}{{.3}^8} + {{\left( {2.3} \right)}^8}{{.2}^2}.5}}\\
 = \dfrac{{{2^{10}}{{.3}^8} - {{2.2}^9}{{.3}^9}}}{{{2^{10}}{{.3}^8} + {2^8}{{.3}^8}{{.2}^2}.5}}\\
 = \dfrac{{{2^{10}}{{.3}^8} - {2^{10}}{{.3}^9}}}{{{2^{10}}{{.3}^8} + {2^{10}}{{.3}^8}.5}}\\
 = \dfrac{{{2^{10}}{{.3}^8}\left( {1 - 3} \right)}}{{{2^{10}}{{.3}^8}.\left( {1 + 5} \right)}}\\
 = \dfrac{{1 - 3}}{{1 + 5}} = \dfrac{{ - 2}}{6} =  - \dfrac{1}{3}\\
3,\\
a,\\
0,4:x = x:0,9\\
 \Leftrightarrow \dfrac{{0,4}}{x} = \dfrac{x}{{0,9}}\\
 \Leftrightarrow 0,4.0,9 = x.x\\
 \Leftrightarrow {x^2} = \dfrac{9}{{25}}\\
 \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{5}\\
x =  - \dfrac{3}{5}
\end{array} \right.\\
b,\\
13\dfrac{1}{3}:1\dfrac{1}{3} = 26:\left( {2x - 1} \right)\\
 \Leftrightarrow \dfrac{{40}}{3}:\dfrac{4}{3} = 26:\left( {2x - 1} \right)\\
 \Leftrightarrow \dfrac{{40}}{3}.\dfrac{3}{4} = 26:\left( {2x - 1} \right)\\
 \Leftrightarrow 10 = 26:\left( {2x - 1} \right)\\
 \Leftrightarrow 2x - 1 = 26:10\\
 \Leftrightarrow 2x - 1 = \dfrac{{13}}{5}\\
 \Leftrightarrow 2x = \dfrac{{13}}{5} + 1\\
 \Leftrightarrow 2x = \dfrac{{18}}{5}\\
 \Leftrightarrow x = \dfrac{{18}}{5}:2\\
 \Leftrightarrow x = \dfrac{9}{5}
\end{array}\)