Đáp án:
$\begin{array}{l}
1)a)x:{\left( { - \dfrac{1}{5}} \right)^3} = {\left( {\dfrac{1}{5}} \right)^2}\\
\Leftrightarrow x = {\left( {\dfrac{1}{5}} \right)^2}.{\left( { - \dfrac{1}{5}} \right)^3}\\
\Leftrightarrow x = - \dfrac{1}{{{5^5}}}\\
Vậy\,x = - \dfrac{1}{{{5^5}}}\\
b)\dfrac{{{{120}^3}}}{{{{40}^3}}}.{x^2} = 2700\\
\Leftrightarrow {\left( {\dfrac{{120}}{{40}}} \right)^3}.{x^2} = 2700\\
\Leftrightarrow {3^3}.{x^2} = 2700\\
\Leftrightarrow {x^2} = 100\\
\Leftrightarrow x = 10;x = - 10\\
Vậy\,x = 10;x = - 10\\
c)\dfrac{x}{y} = \dfrac{5}{4}\\
\Leftrightarrow \dfrac{x}{5} = \dfrac{y}{4} = \dfrac{{y - x}}{{4 - 5}} = \dfrac{4}{{ - 1}} = - 4\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 20\\
y = - 16
\end{array} \right.\\
Vậy\,x = - 20;y = - 16\\
d)2x = 3y \Leftrightarrow \dfrac{x}{3} = \dfrac{y}{2} \Leftrightarrow \dfrac{x}{{21}} = \dfrac{y}{{14}}\\
5y = 7z \Leftrightarrow \dfrac{y}{7} = \dfrac{z}{5} \Leftrightarrow \dfrac{y}{{14}} = \dfrac{z}{{10}}\\
\Leftrightarrow \dfrac{x}{{21}} = \dfrac{y}{{14}} = \dfrac{z}{{10}} = \dfrac{{3x}}{{63}} = \dfrac{{7y}}{{98}} = \dfrac{{5z}}{{50}}\\
= \dfrac{{3x - 7y + 5z}}{{63 - 98 + 50}} = \dfrac{{30}}{{15}} = 2\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 42\\
y = 28\\
z = 20
\end{array} \right.\\
Vậy\,x = 42;y = 28;z = 20\\
e)\left| {2\dfrac{1}{3} - x} \right| - 4\dfrac{1}{3} = 3\dfrac{1}{3}\\
\Leftrightarrow \left| {\dfrac{7}{3} - x} \right| = 3\dfrac{1}{3} + 4\dfrac{1}{3}\\
\Leftrightarrow \left| {\dfrac{7}{3} - x} \right| = \dfrac{{22}}{3}\\
\Leftrightarrow \left[ \begin{array}{l}
\dfrac{7}{3} - x = \dfrac{{22}}{3}\\
\dfrac{7}{3} - x = - \dfrac{{22}}{3}
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{{ - 15}}{3} = - 5\\
x = \dfrac{{29}}{3}
\end{array} \right.\\
Vậy\,x = - 5;x = \dfrac{{29}}{3}\\
C2)\\
{25^{50}} = {\left( {{5^2}} \right)^{50}} = {5^{100}}\\
{2^{300}} = {\left( {{2^3}} \right)^{100}} = {8^{100}} > {5^{100}}\\
Vậy\,{25^{50}} < {2^{300}}
\end{array}$
