\(2,\\ 1,-2+\dfrac{-5}{8}=-\dfrac{21}{8}\\ 2,\dfrac{-2}{5}-\dfrac{-3}{11}=-\dfrac{7}{55}\\ 3,-3\dfrac{1}{2}-2\dfrac{1}{4}=-\dfrac{7}{2}-\dfrac{9}{4}=\dfrac{-23}{4}\\ 4,\dfrac{5}{8}-\left(-\dfrac{2}{5}\right)-\dfrac{3}{10}=\dfrac{29}{40}\\ 5,\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{10}\right)=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{10}=\dfrac{1}{15}\\ 6,\dfrac{1}{12}-\left(-\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4}=\dfrac{1}{2}\\ 7,\dfrac{1}{3}-\left[-\dfrac{5}{4}-\left(\dfrac{1}{4}+\dfrac{3}{8}\right)\right]=\dfrac{1}{3}+\dfrac{5}{4}+\dfrac{5}{8}=\dfrac{53}{24}\\ 8,\dfrac{3}{4}-\left[-\dfrac{5}{3}-\left(\dfrac{1}{12}+\dfrac{2}{9}\right)\right]=\dfrac{3}{4}+\dfrac{5}{3}+\dfrac{11}{36}=\dfrac{49}{18}\)
Bài 4:
9: Ta có: \(\left|x-\dfrac{13}{10}\right|+\dfrac{1}{5}=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\\x-\dfrac{13}{10}=-\dfrac{13}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\\x=0\end{matrix}\right.\)
