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28 tháng 4 2019

a,\(\frac{3}{7}.\frac{4}{9}+\frac{3}{7}.\frac{5}{9}+\frac{5}{14}\)

\(=\frac{3}{7}.\left(\frac{4}{9}+\frac{5}{9}\right)+\frac{5}{14}\)

\(=\frac{3}{7}.1+\frac{5}{14}\)

\(=\frac{3}{7}+\frac{5}{14}=\frac{6}{14}+\frac{5}{14}=\frac{11}{14}\)

b,\(\frac{-11}{23}.\frac{6}{7}+\frac{8}{9}.\frac{-11}{23}-\frac{1}{23}\)

\(=\)\(\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{9}\right)-\frac{1}{23}\)

\(=\frac{-11}{23}.\frac{110}{63}-\frac{1}{23}\)

=\(\frac{-1210}{1449}\)-\(\frac{1}{23}\)

\(=\frac{-1273}{1449}\)

25 tháng 2 2018

a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)

Phần b T2

c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)

25 tháng 2 2018

cảm ơn bạn nha

24 tháng 3 2019

a) A = \(\frac{19}{23}.\frac{-4}{27}-\frac{4}{23}.\frac{2}{7}\)

        = \(\frac{19}{7}.\frac{-4}{23}+\frac{-4}{23}.\frac{2}{7}\)

         = \(\frac{-4}{23}.\left(\frac{19}{7}+\frac{2}{7}\right)\) 

          = \(\frac{-4}{23}.3\)

          = \(\frac{-12}{23}\)

b) B = \(\frac{3}{5}+\frac{2}{5}.\frac{-11}{3}+\frac{2}{3}.\frac{-2}{5}+\frac{14}{15}\)

       = \(\frac{9+14}{15}+\frac{2}{5}.\frac{-11}{3}+\frac{-2}{3}.\frac{2}{5}\)

       = \(\frac{23}{15}+\frac{2}{5}\left(\frac{-11}{3}+\frac{-2}{3}\right)\)

       = \(\frac{23}{15}+\frac{2}{5}.\frac{-13}{3}\)

       = \(\frac{23}{15}+\frac{-26}{15}\)

       = \(\frac{-3}{15}=\frac{-1}{5}\)

12 tháng 2 2017

a/ \(\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(-\frac{9}{10}\right)\)

= \(-\frac{9}{10}.\left(\frac{5}{14}+\frac{1}{7}\right)+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(-\frac{9}{10}.\frac{1}{2}+\frac{1}{10}.\left(-\frac{9}{2}\right)\)

= \(\frac{-9}{20}+\left(-\frac{9}{20}\right)=\frac{-18}{20}=\frac{-9}{10}\)

b/ \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{11}\right).132\)

\(=\left(\frac{1}{2}.132\right)+\left(\frac{1}{3}.132\right)+\left(\frac{1}{4}.132\right)+\left(\frac{1}{6}.132\right)\)\(+\left(\frac{1}{11}.132\right)\)

\(=66+44+33+22+12=177\)

c/ \(-\frac{2}{3}.\left(\frac{8}{9}.\frac{8}{13}-\frac{8}{27}.\frac{8}{13}+\frac{4}{3}.\frac{22}{39}\right)\)

= \(-\frac{2}{3}.\left[\frac{8}{13}\left(\frac{8}{9}-\frac{8}{27}\right)+\frac{88}{117}\right]\)

= \(-\frac{2}{3}.\left(\frac{8}{13}.\frac{16}{27}+\frac{88}{117}\right)\)

= còn lại làm nốt nha! bận ròy

12 tháng 2 2017

gidkjbibvvfrxdrfdfsddf

9 tháng 5 2017

BẠN XEM LẠI CÁI ĐỀ XEM ĐÚNG KO

9 tháng 5 2017

\(\frac{11}{12}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9.\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{11}{12}.\frac{1}{3}=\frac{11}{36}\)

28 tháng 2 2020

                                                    Bài giải

a, \(\frac{7}{12}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{12}\)

\(=\left(\frac{7}{12}-\frac{5}{12}+\frac{5}{6}+\frac{1}{4}\right)-\frac{3}{7}=\left(\frac{7}{12}-\frac{5}{12}+\frac{10}{12}+\frac{3}{12}\right)-\frac{3}{7}=\frac{5}{4}-\frac{3}{7}=\frac{23}{28}\)

b, \(\frac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}=\frac{11\cdot3^{29}-3^{30}}{2^2\cdot3^{28}}=\frac{3^{29}\left(11-3\right)}{3^{28}\cdot4}=\frac{3\cdot8}{4}=6\)

a) Ta có: \(\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{54}{24}\cdot\frac{56}{21}\)

\(=\frac{16}{15}\cdot\frac{-5}{14}\cdot\frac{9}{4}\cdot\frac{8}{3}\)

\(=4\cdot\frac{-1}{3}\cdot\frac{4}{7}\cdot3\)

\(=12\cdot\frac{-4}{21}=\frac{-48}{21}=\frac{-16}{7}\)

b) Ta có: \(5\cdot\frac{7}{5}=\frac{35}{5}=7\)

c) Ta có: \(\frac{1}{7}\cdot\frac{5}{9}+\frac{5}{9}\cdot\frac{1}{7}+\frac{5}{9}\cdot\frac{3}{7}\)

\(=\frac{5}{9}\left(\frac{1}{7}+\frac{1}{7}+\frac{3}{7}\right)\)

\(=\frac{5}{9}\cdot\frac{5}{7}=\frac{25}{63}\)

d) Ta có: \(4\cdot11\cdot\frac{3}{4}\cdot\frac{9}{121}\)

\(=\frac{4\cdot11\cdot3\cdot9}{4\cdot121}=\frac{27}{11}\)

e) Ta có: \(\frac{3}{4}\cdot\frac{16}{9}-\frac{7}{5}:\frac{-21}{20}\)

\(=\frac{4}{3}+\frac{4}{3}=\frac{8}{3}\)

g) Ta có: \(2\frac{1}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\left(\frac{2}{3}+0,4\cdot5\right)\right]\)

\(=\frac{7}{3}-\frac{1}{3}\cdot\left[\frac{-3}{2}+\frac{2}{3}+2\right]\)

\(=\frac{7}{3}-\frac{1}{3}\cdot\frac{7}{6}\)

\(=\frac{7}{3}-\frac{7}{18}=\frac{42}{18}-\frac{7}{18}=\frac{35}{18}\)

14 tháng 7 2020

thank you,very well

30 tháng 4 2019

\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}+\frac{1}{6}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{8}+\frac{1}{8}\cdot\frac{1}{9}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)

30 tháng 4 2019

\(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+...+\frac{1}{8}\cdot\frac{1}{9}\)

\(=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}\)

 \(=\frac{1}{2}-\frac{1}{9}\)

            * LÀM NỐT *

                              #Louis

1 tháng 4 2016

\(\frac{24\cdot47-23}{24+47\cdot23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)

\(=\frac{24\cdot\left(24+23\right)-23}{24+\left(24+23\right)\cdot23}\cdot\frac{3\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}\)

\(=\frac{24^2+24\cdot23-23}{24+24\cdot23+23^2}\cdot\frac{3}{9}\) \(=\frac{24^2+23\cdot\left(24-1\right)}{\left(23+1\right)\cdot24\cdot23^2}\cdot\frac{1}{3}=1\cdot\frac{1}{3}=\frac{1}{3}\)

1 tháng 4 2016

giúp giùm mình đi