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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 = \(\left(11\frac{5}{17}+3\frac{5}{7}\right)-4\frac{5}{17}=11\frac{5}{17}+3\frac{5}{7}-4\frac{5}{17}=\left(11\frac{5}{17}-4\frac{5}{17}\right)+3\frac{5}{7}=7+\frac{26}{7}=\frac{75}{7}\)
1^3-3^5-(-3^5)+1^64-2^9-1^36+1^15
=1+(-3^5+3^5)+1-2^9-1+1
=2-2^9
=-510
\(A=\frac{2}{7}+\frac{-3}{8}+\frac{11}{7}+\frac{1}{3}+\frac{1}{7}+\frac{5}{-8}\)
\(A=\left(\frac{2}{7}+\frac{11}{7}+\frac{1}{7}\right)+\left(\frac{-3}{8}+\frac{5}{-8}\right)+\frac{1}{3}\)
\(A=2-1+\frac{1}{3}\)
\(A=\frac{4}{3}\)
\(B=\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+17\)
\(B=\left(\frac{3}{17}+\frac{14}{17}\right)+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=1+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=18+\frac{-5}{13}+\frac{-18}{35}\)
\(B=\frac{7781}{455}\)
a: \(=5+\dfrac{1}{5}-\dfrac{2}{9}+\dfrac{2}{23}-\dfrac{3}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)
\(=-3+\left(-\dfrac{2}{9}-\dfrac{5}{6}+\dfrac{1}{18}\right)+\left(\dfrac{1}{5}+\dfrac{2}{23}-\dfrac{3}{35}-\dfrac{2}{7}\right)\)
\(=-3+\dfrac{-12-45+1}{18}+\dfrac{7-10-3}{35}+\dfrac{2}{23}\)
\(=3-\dfrac{28}{9}+\dfrac{-6}{35}+\dfrac{2}{23}=\dfrac{-1417}{7245}\)
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)