Đáp án:
$\begin{array}{l}
22)\dfrac{{{{45}^{10}}{{.5}^{10}}}}{{{{75}^{10}}}} = {\left( {\dfrac{{45.5}}{{75}}} \right)^{10}} = {3^{10}}\\
23)\dfrac{{{{\left( {0,8} \right)}^5}}}{{{{\left( {0,4} \right)}^6}}} = {\left( {\dfrac{{0,8}}{{0,4}}} \right)^5}.\dfrac{1}{{0,4}} = {2^5}.\dfrac{5}{2} = 80\\
24)\\
\dfrac{{{2^{15}}{{.9}^4}}}{{{6^3}{{.8}^3}}} = \dfrac{{{2^{15}}{{.3}^8}}}{{{2^3}{{.3}^3}{{.2}^9}}} = {2^3}{.3^5}\\
25)\\
\dfrac{{{8^{10}} + {{30}^{10}}}}{{{8^4} + {4^{11}}}} = \dfrac{{{2^{30}} + {2^{10}}{{.3}^{10}}{{.5}^{10}}}}{{{2^{12}} + {2^{22}}}} = \dfrac{{{2^{10}}\left( {{2^{20}} + {3^{10}}{5^{`10}}} \right)}}{{{2^{12}}\left( {1 + {2^{10}}} \right)}}\\
= \dfrac{{{2^{20}} + {3^{10}}{{.5}^{10}}}}{{{2^2} + {2^{12}}}}\\
Bài3\\
1)x + \dfrac{1}{4} = \dfrac{4}{3}\\
\Leftrightarrow x = \dfrac{{13}}{{12}}\\
Vậyx = \dfrac{{13}}{{12}}\\
2) - x - \dfrac{2}{3} = - \dfrac{6}{7}\\
\Leftrightarrow x = \dfrac{4}{{21}}\\
Vậyx = \dfrac{4}{{21}}\\
3)\dfrac{4}{5} - x = \dfrac{1}{3}\\
\Leftrightarrow x = \dfrac{7}{{15}}\\
Vậy\,x = \dfrac{7}{{15}}\\
5)x + \dfrac{1}{2} = {2^5}:{2^3}\\
\Leftrightarrow x + \dfrac{1}{2} = 4\\
\Leftrightarrow x = \dfrac{7}{2}\\
Vậy\,x = \dfrac{7}{2}\\
6)\dfrac{2}{3} + \dfrac{5}{3}.x = \dfrac{5}{7}\\
\Leftrightarrow \dfrac{5}{3}.x = \dfrac{1}{{21}}\\
\Leftrightarrow x = \dfrac{1}{{35}}\\
Vậy\,x = \dfrac{1}{{35}}\\
7)\left| {x + 5} \right| - 6 = 9\\
\Leftrightarrow \left| {x + 5} \right| = 15\\
\Leftrightarrow \left[ \begin{array}{l}
x + 5 = 15 \Leftrightarrow x = 10\\
x + 5 = - 15 \Leftrightarrow x = - 20
\end{array} \right.\\
Vậy\,x = 10,x = - 20\\
8) - \dfrac{{12}}{{13}}.x - 5 = 6\dfrac{1}{{13}}\\
\Leftrightarrow - \dfrac{{12}}{{13}}.x = \dfrac{{144}}{{13}}\\
\Leftrightarrow x = - 12\\
Vậy\,x = - 12\\
9)\left| {x - 2} \right| = 2\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 2 \Leftrightarrow x = 4\\
x - 2 = - 2 \Leftrightarrow x = 0
\end{array} \right.\\
Vậy\,x = 4,x = 0\\
10)\left| {x + 1} \right| = 2\\
\Leftrightarrow \left[ \begin{array}{l}
x + 1 = 2 \Leftrightarrow x = 1\\
x + 1 = - 2 \Leftrightarrow x = - 3
\end{array} \right.\\
Vậy\,x = 1,x = - 3\\
11)\left| {x - \dfrac{4}{5}} \right| = \dfrac{3}{4}\\
\Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{4}{5} = \dfrac{3}{4}\\
x - \dfrac{4}{5} = - \dfrac{3}{4}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{3}{4} + \dfrac{4}{5} = \dfrac{{31}}{{20}}\\
x = \dfrac{{ - 3}}{4} + \dfrac{4}{5} = \dfrac{1}{{20}}
\end{array} \right.\\
Vậy\,x = \dfrac{{31}}{{20}};x = \dfrac{1}{{20}}\\
12)6 - \left| {\dfrac{1}{2} - x} \right| = \dfrac{2}{5}\\
\Leftrightarrow \left| {\dfrac{1}{2} - x} \right| = 6 - \dfrac{2}{5}\\
\Leftrightarrow \left| {\dfrac{1}{2} - x} \right| = \dfrac{{28}}{5}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{1}{2} - x = \dfrac{{28}}{5}\\
\dfrac{1}{2} - x = \dfrac{{ - 28}}{5}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2} - \dfrac{{28}}{5} = \dfrac{{ - 51}}{{10}}\\
x = \dfrac{1}{2} + \dfrac{{28}}{5} = \dfrac{{61}}{{10}}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 51}}{{10}};x = \dfrac{{61}}{{10}}\\
13){\left( {\dfrac{{ - 2}}{3}} \right)^2}.x = {\left( { - \dfrac{2}{3}} \right)^5}\\
\Leftrightarrow x = {\left( {\dfrac{{ - 2}}{3}} \right)^5}:{\left( {\dfrac{{ - 2}}{3}} \right)^2}\\
\Leftrightarrow x = {\left( {\dfrac{{ - 2}}{3}} \right)^3} = \dfrac{{ - 8}}{{27}}\\
Vậy\,x = \dfrac{{ - 8}}{{27}}\\
14){\left( {\dfrac{{ - 1}}{3}} \right)^3}.x = \dfrac{1}{{81}}\\
\Leftrightarrow x = \dfrac{1}{{81}}:{\left( {\dfrac{{ - 1}}{3}} \right)^3}\\
\Leftrightarrow x = \dfrac{1}{{{3^4}}}:{\left( {\dfrac{{ - 1}}{3}} \right)^3}\\
\Leftrightarrow x = - \dfrac{1}{3}\\
Vậy\,x = \dfrac{{ - 1}}{3}\\
16){2^{x - 1}} = 16\\
\Leftrightarrow {2^{x - 1}} = {2^4}\\
\Leftrightarrow x - 1 = 4\\
\Leftrightarrow x = 5\\
Vậy\,x = 5\\
17){\left( {x - 1} \right)^2} = 25\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 5\\
x - 1 = - 5
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 6\\
x = - 4
\end{array} \right.\\
Vậy\,x = 6;x = - 4
\end{array}$
