Đáp án:

 $a)
 x=\dfrac{-1}{6}\\
b)
 x=\dfrac{-1}{4}\\
c)
x=\dfrac{-15}{49}\\
d)
x=\dfrac{4}{5}\\
e)
x=\dfrac{-1}{2}\\
f)
x=\dfrac{5}{6}\\
g)
 x=\dfrac{89}{4}\\
h)
x=40\\
i)
 {\left[\begin{aligned}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{aligned}\right.}\\
j)
 {\left[\begin{aligned}x=4\\x=0\end{aligned}\right.}\\
k)
{\left[\begin{aligned}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{aligned}\right.}\\
l)
x=\dfrac{-11}{27}$

Giải thích các bước giải:

 $a)
\dfrac{-1}{2}+x=\dfrac{-2}{3}\\
\Leftrightarrow x=\dfrac{-2}{3}+\dfrac{1}{2}\\
\Leftrightarrow x=\dfrac{-4}{6}+\dfrac{3}{6}\\
\Leftrightarrow x=\dfrac{-1}{6}\\
b)
\dfrac{1}{5}+\dfrac{4}{5}:x=-3\\
\Leftrightarrow \dfrac{4}{5}:x=-3-\dfrac{1}{5}\\
\Leftrightarrow \dfrac{4}{5}.\dfrac{1}{x}=\dfrac{-15}{5}-\dfrac{1}{5}\\
\Leftrightarrow \dfrac{4}{5}.\dfrac{1}{x}=\dfrac{-16}{5}\\
\Leftrightarrow \dfrac{1}{x}=\dfrac{-16}{5}.\dfrac{5}{4}=-4\\
\Leftrightarrow x=\dfrac{-1}{4}\\
c)
x:\dfrac{3}{7}+\dfrac{4}{7}=-\dfrac{2}{14}=\dfrac{-1}{7}\\
\Leftrightarrow x:\dfrac{3}{7}=\dfrac{-1}{7}-\dfrac{4}{7}=\dfrac{-5}{7}\\
\Leftrightarrow x=\dfrac{-5}{7}.\dfrac{3}{7}=\dfrac{-15}{49}\\
d)
(x-0,5):\dfrac{30}{100}+\dfrac{1}{3}=1\dfrac{1}{3}\\
\Leftrightarrow (x-\dfrac{1}{2}):\dfrac{3}{10}+\dfrac{1}{3}=\dfrac{4}{3}\\
\Leftrightarrow (x-\dfrac{1}{2}):\dfrac{3}{10}=\dfrac{4}{3}-\dfrac{1}{3}=\dfrac{3}{3}=1\\
\Leftrightarrow x-\dfrac{1}{2}=1.\dfrac{3}{10}=\dfrac{3}{10}\\
\Leftrightarrow x=\dfrac{3}{10}+\dfrac{1}{2}\\
\Leftrightarrow x=\dfrac{3}{10}+\dfrac{5}{10}\\
\Leftrightarrow x=\dfrac{8}{10}=\dfrac{4}{5}\\
e)
\dfrac{1}{3}.\left ( x-1 \right )+\dfrac{1}{2}.\left ( x+2 \right )=0,25\\
\Leftrightarrow \dfrac{1}{3}.x-\dfrac{1}{3} +\dfrac{1}{2}x+1 =\dfrac{1}{4}\\
\Leftrightarrow \left ( \dfrac{1}{3}+\dfrac{1}{2} \right )x=\dfrac{1}{4}+\dfrac{1}{3}-1\\
\Leftrightarrow \left ( \dfrac{2}{6}+\dfrac{3}{6} \right )x=\dfrac{3}{12}+\dfrac{4}{12}-\dfrac{12}{12}\\
\Leftrightarrow \dfrac{5}{6}x=\dfrac{-5}{12}\\
\Leftrightarrow x=\dfrac{-5}{12}.\dfrac{6}{5}=\dfrac{-1}{2}\\
f)
\dfrac{7}{9}:\left ( 2+\dfrac{3}{4}x \right )+\dfrac{5}{9}=\dfrac{23}{27}\\
\Leftrightarrow \dfrac{7}{9}:\left ( 2+\dfrac{3}{4}x \right )=\dfrac{23}{27}-\dfrac{5}{9}\\
\Leftrightarrow \dfrac{7}{9}:\left ( 2+\dfrac{3}{4}x \right )=\dfrac{23}{27}-\dfrac{15}{27}=\dfrac{8}{27}\\
\Leftrightarrow  2+\dfrac{3}{4}x =\dfrac{7}{9}:\dfrac{8}{27}\\
\Leftrightarrow  2+\dfrac{3}{4}x =\dfrac{7}{9}.\dfrac{27}{8}=\dfrac{21}{8}\\
\Leftrightarrow \dfrac{3}{4}x=\dfrac{21}{8}-2\\
\Leftrightarrow \dfrac{3}{4}x=\dfrac{21}{8}-\dfrac{16}{8}=\dfrac{5}{8}\\
\Leftrightarrow x=\dfrac{5}{8}.\dfrac{4}{3}=\dfrac{5}{6}\\
g)
5\dfrac{8}{17}:x+\dfrac{-4}{17}:x=\dfrac{4}{17}\\
\Leftrightarrow \left ( 5\dfrac{8}{17}+\dfrac{-4}{17} \right ):x=\dfrac{4}{17}\\
\Leftrightarrow \left ( \dfrac{93}{17}+\dfrac{-4}{17} \right ):x=\dfrac{4}{17}\\
\Leftrightarrow \dfrac{89}{17}.\dfrac{1}{x}=\dfrac{4}{17}\\
\Leftrightarrow \dfrac{1}{x}=\dfrac{4}{17}.\dfrac{17}{89}=\dfrac{4}{89}\\
\Leftrightarrow x=\dfrac{89}{4}\\
h)
x-\dfrac{1}{4}x=60-x+\dfrac{1}{4}x\\
\Leftrightarrow x-\dfrac{1}{4}x+x-\dfrac{1}{4}x=60\\
\Leftrightarrow 2x-\dfrac{2}{4}x=60\\
\Leftrightarrow \left ( 2-\dfrac{1}{2} \right )x=60\\
\Leftrightarrow \left ( \dfrac{4}{2}-\dfrac{1}{2} \right )x=60\\
\Leftrightarrow \dfrac{3}{2}x=60\\
\Leftrightarrow x=60.\dfrac{2}{3}=40\\
i)
\left ( x+\dfrac{1}{2} \right ).\left ( \dfrac{2}{3}-2x \right )=0\\
\Leftrightarrow {\left[\begin{aligned}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=-\dfrac{1}{2}\\2x=\dfrac{2}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{aligned}\right.}\\
j)
|x-2|-1\dfrac{2}{7}=\dfrac{5}{7}\\
\Leftrightarrow |x-2|=\dfrac{5}{7}+\dfrac{9}{7}=\dfrac{14}{7}=2\\
\Leftrightarrow {\left[\begin{aligned}x-2=2\\x-2=-2\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=4\\x=0\end{aligned}\right.}\\
k)
\left ( x+\dfrac{1}{5} \right )^2+\dfrac{17}{25}=\dfrac{26}{25}\\
\Leftrightarrow \left ( x+\dfrac{1}{5} \right )^2=\dfrac{26}{25}-\dfrac{17}{25}=\dfrac{9}{25}\\
\Leftrightarrow {\left[\begin{aligned}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{3}{5}-\dfrac{1}{5}\\x=-\dfrac{3}{5}-\dfrac{1}{5}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{aligned}\right.}\\
l)
-1-\left ( 3x-\dfrac{7}{9} \right )^3=7\\
\Leftrightarrow \left ( 3x-\dfrac{7}{9} \right )^3=-1-7=-8\\
\Leftrightarrow 3x-\dfrac{7}{9}=-2\\
\Leftrightarrow 3x=-2+\dfrac{7}{9}\\
\Leftrightarrow 3x=\dfrac{-18}{9}+\dfrac{7}{9}\\
\Leftrightarrow 3x=\dfrac{-11}{9}\\
\Leftrightarrow x=\dfrac{-11}{27}$