Đáp án:

$\begin{array}{l}
a)x:4\dfrac{1}{3} =  - 2,5\\
 \Rightarrow x:\dfrac{{13}}{3} =  - \dfrac{5}{2}\\
 \Rightarrow x = \dfrac{{ - 5}}{2}.\dfrac{3}{{13}}\\
 \Rightarrow x = \dfrac{{ - 15}}{{26}}\\
Vậy\,x = \dfrac{{ - 15}}{{26}}\\
b)\dfrac{3}{4} + \dfrac{1}{4}:x = 1\\
 \Rightarrow \dfrac{1}{4}:x = 1 - \dfrac{3}{4}\\
 \Rightarrow \dfrac{1}{4}:x = \dfrac{1}{4}\\
 \Rightarrow x = 1\\
Vậy\,x = 1\\
c)\dfrac{{ - 2}}{3}.x + \dfrac{1}{5} = \dfrac{3}{{10}}\\
 \Rightarrow  - \dfrac{2}{3}.x = \dfrac{3}{{10}} - \dfrac{1}{5}\\
 \Rightarrow  - \dfrac{2}{3}.x = \dfrac{1}{{10}}\\
 \Rightarrow x = \dfrac{1}{{10}}.\dfrac{{ - 3}}{2}\\
 \Rightarrow x = \dfrac{{ - 3}}{{20}}\\
Vậy\,x = \dfrac{{ - 3}}{{20}}\\
d)\left| x \right| - \dfrac{3}{4} = \dfrac{5}{3}\\
 \Rightarrow \left| x \right| = \dfrac{5}{3} + \dfrac{3}{4}\\
 \Rightarrow \left| x \right| = \dfrac{{29}}{{12}}\\
 \Rightarrow x =  \pm \dfrac{{29}}{{12}}\\
Vậy\,x =  \pm \dfrac{{29}}{{12}}\\
e)\dfrac{{17}}{2} - \left| {2x - \dfrac{3}{4}} \right| = \dfrac{{ - 7}}{4}\\
 \Rightarrow \left| {2x - \dfrac{3}{4}} \right| = \dfrac{{17}}{2} + \dfrac{7}{4}\\
 \Rightarrow \left| {2x - \dfrac{3}{4}} \right| = \dfrac{{41}}{4}\\
 \Rightarrow \left[ \begin{array}{l}
2x - \dfrac{3}{4} = \dfrac{{41}}{4}\\
2x - \dfrac{3}{4} =  - \dfrac{{41}}{4}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
2x = 11\\
2x = \dfrac{{ - 19}}{2}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{{11}}{2}\\
x = \dfrac{{ - 19}}{4}
\end{array} \right.\\
Vậy\,x = \dfrac{{ - 19}}{4};x = \dfrac{{11}}{2}\\
f)\left| {2x - \dfrac{1}{3}} \right| + \dfrac{5}{6} = 1\\
 \Rightarrow \left| {2x - \dfrac{1}{3}} \right| = 1 - \dfrac{5}{6} = \dfrac{1}{6}\\
 \Rightarrow \left[ \begin{array}{l}
2x - \dfrac{1}{3} = \dfrac{1}{6}\\
2x - \dfrac{1}{3} =  - \dfrac{1}{6}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
2x = \dfrac{1}{6} + \dfrac{1}{3} = \dfrac{1}{2}\\
2x = \dfrac{1}{6}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = \dfrac{1}{{12}}
\end{array} \right.
\end{array}$

$\begin{array}{l}
g){\left( {x + \dfrac{1}{5}} \right)^2} + \dfrac{{17}}{{25}} = \dfrac{{26}}{{25}}\\
 \Rightarrow {\left( {x + \dfrac{1}{5}} \right)^2} = \dfrac{9}{{25}}\\
 \Rightarrow \left[ \begin{array}{l}
x + \dfrac{1}{5} = \dfrac{3}{5}\\
x + \dfrac{1}{5} =  - \dfrac{3}{5}
\end{array} \right.\\
 \Rightarrow \left[ \begin{array}{l}
x = \dfrac{2}{5}\\
x =  - \dfrac{4}{5}
\end{array} \right.\\
h) - 1\dfrac{5}{{27}} - {\left( {3x - \dfrac{7}{9}} \right)^3} = \dfrac{{ - 24}}{{27}}\\
 \Rightarrow \dfrac{{ - 32}}{{27}} - {\left( {3x - \dfrac{7}{9}} \right)^3} = \dfrac{{ - 24}}{{27}}\\
 \Rightarrow {\left( {3x - \dfrac{7}{9}} \right)^3} = \dfrac{{ - 8}}{{27}}\\
 \Rightarrow 3x - \dfrac{7}{9} = \dfrac{{ - 2}}{3}\\
 \Rightarrow 3x = \dfrac{{ - 2}}{3} + \dfrac{7}{9}\\
 \Rightarrow 3x = \dfrac{1}{9}\\
 \Rightarrow x = \dfrac{1}{{27}}\\
Vậy\,x = \dfrac{1}{{27}}\\
i)\left( {x + \dfrac{1}{2}} \right)\left( {\dfrac{2}{3} - 2x} \right) = 0\\
 \Rightarrow \left[ \begin{array}{l}
x =  - \dfrac{1}{2}\\
x = \dfrac{1}{3}
\end{array} \right.\\
Vậy\,x =  - \dfrac{1}{2};x = \dfrac{1}{3}\\
k)x + 25\% .x = 125\\
 \Rightarrow x + 0,25.x = 125\\
 \Rightarrow 1,25.x = 125\\
 \Rightarrow x = 100\\
Vậy\,x = 100\\
l)\dfrac{2}{3}.x - \dfrac{1}{2}.x = \dfrac{5}{{12}}\\
 \Rightarrow \left( {\dfrac{2}{3} - \dfrac{1}{2}} \right).x = \dfrac{5}{{12}}\\
 \Rightarrow \dfrac{1}{6}.x = \dfrac{5}{{12}}\\
 \Rightarrow x = \dfrac{5}{2}\\
Vậy\,x = \dfrac{5}{2}
\end{array}$