Đáp án:
$\begin{array}{l}
a)\left| { - 2x + \dfrac{1}{3}} \right| - \dfrac{1}{3} = 2\dfrac{2}{3}\\
\Leftrightarrow \left| {2x - \dfrac{1}{3}} \right| = 2\dfrac{2}{3} + \dfrac{1}{3}\\
\Leftrightarrow \left| {2x - \dfrac{1}{3}} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{3} = 3\\
2x - \dfrac{1}{3} = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{5}{3}\\
x = \dfrac{{ - 4}}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{5}{3};x = - \dfrac{4}{3}\\
b)\dfrac{{x - 1}}{{x - 5}} = \dfrac{6}{7}\\
\Leftrightarrow 7.\left( {x - 1} \right) = 6.\left( {x - 5} \right)\\
\Leftrightarrow 7x - 7 = 6x - 30\\
\Leftrightarrow x = - 23\\
Vậy\,x = - 23\\
c)\dfrac{x}{{15}} = \dfrac{y}{{10}} = \dfrac{z}{6} = \dfrac{{x + y - z}}{{15 + 10 - 6}} = \dfrac{{95}}{{19}} = 5\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 5.15 = 75\\
y = 5.10 = 50\\
z = 5.6 = 30
\end{array} \right.\\
Vậy\,x = 75;y = 50;z = 30\\
d)5x = 8y = 20z\\
\Leftrightarrow \dfrac{{5x}}{{40}} = \dfrac{{8y}}{{40}} = \dfrac{{20z}}{{40}}\\
= \dfrac{x}{8} = \dfrac{y}{5} = \dfrac{z}{2} = \dfrac{{x - y - z}}{{8 - 5 - 2}} = \dfrac{3}{1} = 3\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 3.8 = 24\\
y = 3.5 = 15\\
z = 3.2 = 6
\end{array} \right.\\
Vậy\,x = 24;y = 15;z = 6\\
e)2x = 3y \Leftrightarrow \dfrac{x}{3} = \dfrac{y}{2} = \dfrac{x}{{21}} = \dfrac{y}{{14}}\\
5y = 7z \Leftrightarrow \dfrac{y}{7} = \dfrac{z}{5} = \dfrac{y}{{14}} = \dfrac{z}{{10}}\\
\Leftrightarrow \dfrac{x}{{21}} = \dfrac{y}{{14}} = \dfrac{z}{{10}} = \dfrac{{3x}}{{63}} = \dfrac{{5z}}{{50}} = \dfrac{{7y}}{{108}}\\
= \dfrac{{3x - 5z + 7y}}{{63 - 50 + 108}} = \dfrac{{30}}{5} = 6\\
\Leftrightarrow \left\{ \begin{array}{l}
x = 126\\
y = 6.14 = 84\\
z = 6.10 = 60
\end{array} \right.\\
Vậy\,x = 126;y = 84;z = 60
\end{array}$
