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\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
a) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2-\frac{11}{5}:\frac{-11}{5}=\left(-\frac{1}{10}\right)^2+1=1\frac{1}{100}\)
b) \(\left(-\frac{5}{7}\right)^2+8.\left(0,5\right)^2+\left(-1\right)^{2010}=\frac{25}{49}+2+1=3\frac{25}{49}\)
c) \(\frac{9999^2}{3333^2}+\left(0,5\right)^2.\left(-2\right)^4-\left(-\frac{4}{3}\right)^2=9+1-\frac{16}{9}=8\frac{2}{9}\)
d) \(\left|-\frac{2}{5}+\frac{1}{7}\right|:\frac{-3}{35}+\frac{-3}{7}.\frac{7}{5}=\frac{9}{35}.\frac{35}{-3}-\frac{3}{5}=-3\frac{3}{5}\)
e) \(\frac{1}{2}-\left(-0,4\right)+\frac{1}{3}+\frac{1}{5}-\frac{-1}{6}+\frac{-4}{35}+\frac{1}{41}\)
\(=\frac{1}{2}+\frac{2}{5}+\frac{1}{3}+\frac{1}{5}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}=1\frac{732}{1435}\)
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)
a) \(-\frac{4}{7}+\frac{\left(-5\right).\left(-39\right)}{13.25}+\frac{\left(-1\right).6}{42.\left(-5\right)}=-\frac{4}{7}+\frac{\left(-1\right).3}{1.5}+\frac{\left(-1\right).1}{7.\left(-5\right)}=-\frac{4}{7}+\frac{3}{5}+\frac{1}{35}\)
\(=-\frac{20}{35}+\frac{21}{35}+\frac{1}{35}=\frac{2}{35}\)
b) \(=\frac{2}{9}.\left[-\frac{4}{45}:\left(\frac{3}{15}-\frac{2}{15}\right)+1\frac{2}{3}\right]+\frac{5}{27}=\frac{2}{9}.\left[-\frac{4}{45}:\frac{1}{15}+1\frac{2}{3}\right]+\frac{5}{27}\)
\(=\frac{2}{9}.\left[\frac{\left(-4\right).15}{45.1}+1\frac{2}{3}\right]+\frac{5}{27}==\frac{2}{9}.\left[\frac{\left(-4\right).1}{3.1}+1\frac{2}{3}\right]+\frac{5}{27}\)
\(==\frac{2}{9}.\left[-\frac{4}{3}+\frac{5}{3}\right]+\frac{5}{27}=\frac{2.1}{9.3}+\frac{5}{27}=\frac{2}{27}+\frac{5}{27}=\frac{7}{27}\)