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a) A = \(9\frac{3}{8}-\left(2\frac{3}{5}+2\frac{3}{8}\right)=9\frac{3}{8}-2\frac{3}{5}-2\frac{3}{8}=\left(9\frac{3}{8}-2\frac{3}{8}\right)-2\frac{3}{5}=7-\frac{13}{5}=\frac{22}{5}\)
b) B = \(\left(15\frac{3}{5}+5\frac{3}{4}\right)-8\frac{3}{5}=15\frac{3}{5}+5\frac{3}{4}-8\frac{3}{5}=\left(15\frac{3}{5}-8\frac{3}{5}\right)+5\frac{3}{4}=7+\frac{23}{4}=\frac{51}{4}\)
c) C = \(17\frac{1}{4}-\left(2\frac{3}{7}+7\frac{1}{4}\right)=17\frac{1}{4}-2\frac{3}{7}-7\frac{1}{4}=\left(17\frac{1}{4}-7\frac{1}{4}\right)-2\frac{3}{7}=10-\frac{17}{7}=\frac{53}{7}\)
d) D = \(\left(11\frac{5}{17}+3\frac{5}{7}\right)-4\frac{5}{17}=11\frac{5}{17}+3\frac{5}{7}-4\frac{5}{17}=\left(11\frac{5}{17}-4\frac{5}{17}\right)+3\frac{5}{7}=7+\frac{26}{7}=\frac{75}{7}\)
a) Ta có: \(\frac{7}{8}+\frac{15}{23}+\frac{1}{8}+\left(-\frac{15}{23}\right)+\frac{1}{3}\)
\(=\left(\frac{7}{8}+\frac{1}{8}\right)+\left(\frac{15}{23}-\frac{15}{23}\right)+\frac{1}{3}\)
\(=1+\frac{1}{3}=\frac{4}{3}\)
b) Ta có: \(\frac{3-\frac{2}{3}}{\frac{1}{3}-3}\)
\(=\left(3-\frac{2}{3}\right)\cdot\frac{1}{\frac{1}{3}-3}=\left(3-\frac{2}{3}\right)\cdot\frac{1}{-\frac{8}{3}}\)
\(=\left(3-\frac{2}{3}\right)\cdot\frac{-3}{8}\)
\(=-\frac{9}{8}+\frac{1}{4}=\frac{-9}{8}+\frac{2}{8}=-\frac{7}{8}\)
c) Ta có: \(8\cdot\left(-\frac{1}{2}\right)^3+3\cdot\left(-\frac{1}{2}\right)^2-0,75\)
\(=8\cdot\frac{-1}{8}+3\cdot\frac{1}{4}-\frac{3}{4}\)
\(=-1\)