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a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)
\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)
\(=10:\left(\frac{-2}{3}\right)\)
\(=-15\)
b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\frac{-8}{27}+\frac{1}{10}\)
\(=\frac{-8}{3}+\frac{1}{10}\)
\(=\frac{-77}{30}\)
c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)
\(=\frac{-1}{3}\)
\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)
\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)
\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)
\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)
\(=-15\)
\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)
\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)
\(=-\frac{8}{3}+\frac{1}{10}\)
\(=-\frac{77}{30}\)
\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)
\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)
\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)
\(=\frac{1}{3}\)
\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
