8) Ta có: \(x+\dfrac{3}{2}=-\dfrac{5}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{2}=\dfrac{-10}{6}-\dfrac{9}{6}\)
hay \(x=-\dfrac{19}{6}\)
Vậy: \(x=-\dfrac{19}{6}\)
10) Ta có: \(\left|x-\dfrac{1}{2}\right|+75\%=\dfrac{9}{10}\)
\(\Leftrightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{3}{20}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{20}\\x-\dfrac{1}{2}=-\dfrac{3}{20}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{20}\\x=\dfrac{7}{20}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{13}{20};\dfrac{7}{20}\right\}\)
11) Ta có: \(x+\dfrac{2}{3}=-\dfrac{1}{2}\)
nên \(x=-\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{-3}{6}-\dfrac{4}{6}\)
hay \(x=-\dfrac{7}{6}\)
Vậy: \(S=\left\{-\dfrac{7}{6}\right\}\)