Đáp án:

 $1)
x=\dfrac{-23}{6}\\
2)
 x=\dfrac{-10}{9}\\
3)
 x=\dfrac{-1}{5}\\
4)
{\left[\begin{aligned}x=\dfrac{5}{4}\\ x=-\dfrac{1}{4}\end{aligned}\right.}\\
5)
 {\left[\begin{aligned}x=\dfrac{5}{2}\\x=2\end{aligned}\right.}\\
6)
 {\left[\begin{aligned}x = \dfrac{7}{6}\\ x = \dfrac{-1}{6}\end{aligned}\right.}\\
7)
{\left[\begin{aligned}x=\dfrac{4}{5}\\x=\dfrac{-11}{5}\end{aligned}\right.}\\
8)
 {\left[\begin{aligned}x =-\dfrac{1}{3}\\x=\dfrac{3}{8}\end{aligned}\right.}\\
9)
x\dfrac{-5}{2}\\$

Giải thích các bước giải:

 $1)
-1\dfrac{1}{3}-x=2,5\\
\Leftrightarrow x=\dfrac{-4}{3}-\dfrac{5}{2}\\
\Leftrightarrow x=\dfrac{-8}{6}-\dfrac{15}{6}=\dfrac{-23}{6}\\
2)
\dfrac{1}{3}:x-\dfrac{20}{100}=\dfrac{16}{-32}\\
\Leftrightarrow \dfrac{1}{3}.\dfrac{1}{x}-\dfrac{1}{5}=\dfrac{-1}{2}\\
\Leftrightarrow \dfrac{1}{3}.\dfrac{1}{x}=\dfrac{-1}{2}+\dfrac{1}{5}\\
\Leftrightarrow \dfrac{1}{3}.\dfrac{1}{x}=\dfrac{-5}{10}+\dfrac{2}{10}=\dfrac{-3}{10}\\
\Leftrightarrow \dfrac{1}{x}=\dfrac{-3}{10}.3=\dfrac{-9}{10}\\
\Leftrightarrow x=\dfrac{-10}{9}\\
3)
\dfrac{2}{3}+\dfrac{1}{3}:x=-1\\
\Leftrightarrow \dfrac{1}{3}:x=-1-\dfrac{2}{3}\\
\Leftrightarrow \dfrac{1}{3}.\dfrac{1}{x}=\dfrac{-3}{3}-\dfrac{2}{3}\\
\Leftrightarrow \dfrac{1}{3}.\dfrac{1}{x}=\dfrac{-5}{3}\\
\Leftrightarrow \dfrac{1}{x}=\dfrac{-5}{3}.3=-5\\
\Leftrightarrow x=\dfrac{-1}{5}\\
4)
1-\left | x-\dfrac{1}{2} \right |=\dfrac{1}{4}\\
\Leftrightarrow \left | x-\dfrac{1}{2} \right |=1-\dfrac{1}{4}\\
\Leftrightarrow \left | x-\dfrac{1}{2} \right |=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{1}{2}=\dfrac{3}{4}\\ x-\dfrac{1}{2}=-\dfrac{3}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{3}{4}+\dfrac{1}{2}\\ x=-\dfrac{3}{4}+\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{3}{4}+\dfrac{2}{4}\\ x=-\dfrac{3}{4}+\dfrac{2}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{5}{4}\\ x=-\dfrac{1}{4}\end{aligned}\right.}\\
5)
\left | \dfrac{2}{3}x-1\dfrac{1}{2} \right |=\dfrac{1}{6}\\
\Leftrightarrow \left | \dfrac{2}{3}x-\dfrac{3}{2} \right |=\dfrac{1}{6}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{2}{3}x-\dfrac{3}{2}=\dfrac{1}{6}\\\dfrac{2}{3}x-\dfrac{3}{2}=-\dfrac{1}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{3}{2}\\\dfrac{2}{3}x=-\dfrac{1}{6}+\dfrac{3}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{2}{3}x=\dfrac{1}{6}+\dfrac{9}{6}\\\dfrac{2}{3}x=-\dfrac{1}{6}+\dfrac{9}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}\dfrac{2}{3}x=\dfrac{10}{6}\\\dfrac{2}{3}x=\dfrac{8}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{10}{6}.\dfrac{3}{2}\\x=\dfrac{8}{6}.\dfrac{3}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{5}{2}\\x=2\end{aligned}\right.}\\
6)
\left ( x-\dfrac{1}{2} \right )^2=\dfrac{4}{9}\\
\Leftrightarrow  x-\dfrac{1}{2} =\pm \dfrac{2}{3}\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{1}{2} = \dfrac{2}{3}\\ x-\dfrac{1}{2} =- \dfrac{2}{3}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x = \dfrac{2}{3}+\dfrac{1}{2}\\ x =- \dfrac{2}{3}+\dfrac{1}{2}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x = \dfrac{4}{6}+\dfrac{3}{6}\\ x =- \dfrac{4}{6}+\dfrac{3}{6}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x = \dfrac{7}{6}\\ x = \dfrac{-1}{6}\end{aligned}\right.}\\
7)
\left ( x-\dfrac{4}{5} \right ).\left ( x+2\dfrac{1}{5} \right )=0\\
\Leftrightarrow \left ( x-\dfrac{4}{5} \right ).\left ( x+\dfrac{11}{5} \right )=0\\
\Leftrightarrow {\left[\begin{aligned}x-\dfrac{4}{5}=0\\x+\dfrac{11}{5}=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x=\dfrac{4}{5}\\x=\dfrac{-11}{5}\end{aligned}\right.}\\
8)
\left ( x+\dfrac{1}{3} \right ).\left ( \dfrac{3}{4}-2x \right )=0\\
\Leftrightarrow {\left[\begin{aligned}x+\dfrac{1}{3} =0\\\dfrac{3}{4}-2x=0\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =-\dfrac{1}{3}\\2x=\dfrac{3}{4}\end{aligned}\right.}\\
\Leftrightarrow {\left[\begin{aligned}x =-\dfrac{1}{3}\\x=\dfrac{3}{8}\end{aligned}\right.}\\
9)
\dfrac{-4}{5}x+1\dfrac{1}{3}x=-1\dfrac{1}{3}\\
\Leftrightarrow \dfrac{-4}{5}x+\dfrac{4}{3}x=\dfrac{-4}{3}\\
\Leftrightarrow \left (\dfrac{-4}{5}+\dfrac{4}{3}  \right )x=\dfrac{-4}{3}\\
\Leftrightarrow \left (\dfrac{-12}{15}+\dfrac{20}{15}  \right )x=\dfrac{-4}{3}\\
\Leftrightarrow \dfrac{8}{15}x=\dfrac{-4}{3}\\
\Leftrightarrow x=\dfrac{-4}{3}.\dfrac{15}{8}=\dfrac{-5}{2}\\$