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19 tháng 7 2018

A=\([\)\(\frac{2}{7}\)\(\times\)(\(\frac{1}{4}-\frac{1}{3}\))\(]\)\(\div\)\([\)(\(\frac{2}{7}\times\)(\(\frac{3}{9}-\frac{2}{5}\))\(]\)
  =(\(\frac{2}{7}\times\)\(\frac{-1}{12}\))\(\div(\)\(\frac{2}{7}\times\)\(\frac{-1}{15}\))
=\(\frac{-1}{42}\)\(\div\)\(\frac{-2}{35}\)
=\(\frac{-1}{42}\)\(\times\)\(\frac{35}{-2}\)
=\(\frac{5}{12}\)

a) \(\frac{-1}{2}+\frac{-1}{9}-\frac{-3}{5}+\frac{1}{2006}-\frac{-2}{7}-\frac{7}{18}+\frac{4}{35}\)

\(=\left(\frac{-1}{2}-\frac{1}{9}-\frac{7}{18}\right)+\left(\frac{3}{5}+\frac{4}{35}\right)+\frac{1}{2006}\)

\(=\left(\frac{-9}{18}-\frac{2}{18}-\frac{7}{18}\right)+\left(\frac{21}{35}+\frac{4}{35}\right)+\frac{1}{2006}\)

\(=\left(\frac{-9-2-7}{18}\right)+\left(\frac{21+4}{35}\right)+\frac{1}{2006}\)

\(=\left(\frac{-18}{18}\right)+\left(\frac{25}{35}\right)+\frac{1}{2006}\)

\(=\left(-1\right)+\frac{5}{7}+\frac{1}{2006}\)\(=\frac{-4005}{14042}\)

b) \(\frac{1}{3}-\frac{3}{4}+\frac{3}{5}+\frac{1}{2007}-\frac{1}{36}+\frac{1}{15}-\frac{2}{9}\)

\(=\left(\frac{1}{3}+\frac{1}{2007}-\frac{2}{9}\right)-\left(\frac{3}{4}+\frac{1}{36}\right)+\left(\frac{3}{5}+\frac{1}{15}\right)\)

\(=\left(\frac{669}{2007}+\frac{1}{2007}-\frac{446}{2007}\right)-\left(\frac{27}{36}+\frac{1}{36}\right)+\left(\frac{9}{15}+\frac{1}{15}\right)\)

\(=\frac{224}{2007}-\frac{28}{36}+\frac{10}{15}\)

\(=\frac{224}{2007}-\frac{1561}{2007}+\frac{1338}{2007}\)\(=\frac{1}{2007}\)

Bài 1:

a) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)

\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)

\(=4\frac{5}{7}-1\frac{3}{4}\)

\(=\frac{33}{7}-\frac{7}{4}\)

\(=\frac{132}{28}-\frac{49}{28}=\frac{83}{28}\)

b) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)

\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)

\(=4\frac{5}{9}-2\frac{3}{4}\)

\(=\frac{41}{9}-\frac{11}{4}\)

\(=\frac{164}{36}-\frac{99}{36}=\frac{65}{36}\)

c) Ta có: \(\frac{-3}{5}\cdot\frac{5}{7}+\frac{-3}{5}\cdot\frac{3}{7}+\frac{-3}{5}\cdot\frac{6}{7}\)

\(=\frac{-3}{5}\cdot\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)\)

\(=\frac{-3}{5}\cdot2=-\frac{6}{5}\)

d) Ta có: \(\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{4}{3}\)

\(=\frac{1}{3}\cdot\frac{4}{5}+\frac{1}{3}\cdot\frac{6}{5}-\frac{1}{3}\cdot4\)

\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}-4\right)\)

\(=\frac{1}{3}\cdot\left(-2\right)=\frac{-2}{3}\)

17 tháng 6 2015

mh biết làm bài này rùi bn có cần mi2h đang cho bn ko?

11 tháng 7 2016

c.\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)

\(\frac{\frac{25}{108}.\frac{5751}{25}+\frac{187}{4}}{\frac{100}{21}:-\frac{41}{21}}\)

\(\frac{\frac{213}{4}+\frac{187}{4}}{-\frac{100}{41}}\)

\(\frac{100}{-\frac{100}{41}}=-41\)

11 tháng 7 2016

a. \(\frac{4}{9}:-\frac{1}{7}+6\frac{5}{9}:-\frac{1}{7}\)

\(\left(\frac{4}{9}+6\frac{5}{9}\right):-\frac{1}{7}\)

\(7:-\frac{1}{7}=-49\)