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\(\left(5-\frac{43}{10}\right)-\left(\frac{42}{19}-\left(\frac{7}{2}-\frac{59}{10}\right)\right)+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)
\(=5-\frac{43}{10}-\left(\frac{42}{19}-\frac{7}{2}+\frac{59}{10}\right)+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)
\(=5-\frac{43}{10}-\frac{42}{19}+\frac{7}{2}-\frac{59}{10}+\frac{42}{19}+\frac{59}{10}+\frac{4}{5}\)
\(=5-\frac{43}{10}+\frac{7}{2}+\frac{4}{5}\)
\(=5+\frac{35}{10}+\frac{8}{10}-\frac{43}{10}=5\)
\(=\left[\frac{3}{5}-\frac{1}{35}-\left(\frac{-3}{7}\right)\right]+\left[\frac{3}{11}-\frac{3}{4}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{21}{35}-\frac{1}{35}+\frac{15}{35}\right]+\left[\frac{12}{44}-\frac{33}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=\left[\frac{20}{35}+\frac{15}{35}\right]+\left[\frac{-21}{44}+\left(\frac{-23}{44}\right)\right]\)
\(=1+\left(-1\right)\)
\(=0\)
a)\(\left( { - 0,4} \right) + \frac{3}{8} + \left( { - 0,6} \right) = \left[ {\left( { - 0,4} \right) + \left( { - 0,6} \right)} \right] + \frac{3}{8} = - 1 + \frac{3}{8} = \frac{{ - 5}}{8}\).
b)
\(\frac{4}{5} - 1,8 + 0,375 + \frac{5}{8} = (0,8 - 1,8) + (0,375 + 0,625) = ( - 1) + 1 = 0\)
\(=\left(-\frac{1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{2}{7}+\frac{4}{35}\right)+\frac{1}{127}\)
\(=\left(-\frac{9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{10}{35}+\frac{4}{35}\right)+\frac{1}{127}\)
\(=\left(-\frac{18}{18}\right)+\frac{35}{35}+\frac{1}{127}\)
\(=-1+1+\frac{1}{127}\)
\(=\frac{1}{127}\)
a. \(\frac{11}{19}.\frac{13}{7}+\frac{13}{7}.\frac{8}{19}=\frac{13}{7}\left(\frac{11}{19}+\frac{8}{19}\right)=\frac{13}{7}.1=\frac{13}{7}\)
b. \(\left(\frac{1}{3}\right)^2-\left(\frac{3}{4}\right)^3.\left(\frac{4}{3}\right)^3=\frac{1}{9}-\left(\frac{3}{4}.\frac{4}{3}\right)^3=\frac{1}{9}-1=-\frac{8}{9}\)
c. \(\left|1-\frac{2}{3}\right|-2.\left(\frac{-209}{2009}\right)^0=\frac{1}{3}-2.1=\frac{1}{3}-2=-\frac{5}{3}\)
\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)
\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}-\frac{23}{44}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\frac{35}{35}+\left(\frac{-44}{44}\right)+\frac{2}{97}=1+\left(-1\right)+\frac{2}{97}=\frac{2}{97}\)
\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}-\frac{23}{44}\right)+\frac{2}{97}\)
\(=-1+1+\frac{2}{97}\)
\(=\frac{2}{97}\)
\(\left(\frac{3}{7}-\frac{5}{2}-\frac{3}{5}\right)-\left(-\frac{4}{7}-\frac{3}{2}+\frac{2}{5}\right)=\frac{3}{7}-\frac{5}{2}-\frac{3}{5}+\frac{4}{7}+\frac{3}{2}-\frac{2}{5}\)
\(=\left(\frac{3}{7}+\frac{4}{7}\right)+\left(-\frac{5}{2}+\frac{3}{2}\right)+\left(-\frac{3}{5}-\frac{2}{5}\right)\)
\(=1+\left(-1\right)+\left(-1\right)\)
\(=-1\)
Hok tốt nha^^
\(\left(\frac{42}{32}:\frac{7}{8}\right).\frac{3}{4}\)
\(=\left(\frac{42}{32}.\frac{8}{7}\right).\frac{3}{4}\)
\(=\frac{3}{2}.\frac{3}{4}\)
\(=\frac{9}{8}\)