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11 tháng 9 2017

a) \(1\frac{1}{3}+2\frac{1}{2}\)\(\frac{4}{3}+\frac{5}{2}=\frac{8}{6}+\frac{15}{6}=\frac{23}{6}\)

b) \(3\frac{2}{5}-1\frac{1}{10}=\frac{17}{5}-\frac{11}{10}=\frac{34}{10}-\frac{11}{10}=\frac{23}{10}\)

c) \(2+\frac{5}{7}=\frac{14}{7}+\frac{5}{7}=\frac{19}{7}\)

11 tháng 9 2017

a) \(1\frac{1}{3}+2\frac{1}{2}\)

\(=\frac{4}{3}+\frac{5}{2}\)

\(=\frac{8}{6}+\frac{15}{6}\)

\(=\frac{23}{6}\)

b) \(3\frac{2}{5}-1\frac{1}{10}\)

\(=\frac{17}{5}-\frac{11}{10}\)

\(=\frac{34}{10}-\frac{11}{10}\)

\(=\frac{23}{10}\)

c) \(2+\frac{5}{7}\)

\(=\frac{14}{7}+\frac{5}{7}\)

\(=\frac{19}{7}\)

18 tháng 6 2017

a,  \(A=\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)

\(A=\left(\frac{2}{5}-\frac{3}{4}\right)+\left(\frac{-1}{6}-\frac{-2}{3}\right)\)

\(A=\left(\frac{8}{20}-\frac{15}{20}\right)+\left(\frac{-3}{18}-\frac{-12}{18}\right)\)

\(A=\frac{-7}{20}+\frac{1}{2}\)

\(\Rightarrow A=\frac{-7}{20}+\frac{10}{20}=\frac{3}{20}\)

18 tháng 6 2017

b, \(B=\frac{7}{10}-\frac{-3}{4}+\frac{-5}{6}-\frac{1}{5}+\frac{-2}{3}\)

\(B=\left(\frac{7}{10}-\frac{1}{5}\right)+\left(\frac{-5}{6}+\frac{-2}{3}\right)-\frac{-3}{4}\)

\(B=\left(\frac{7}{10}-\frac{2}{10}\right)+\left(\frac{-5}{6}+\frac{-4}{6}\right)-\frac{-3}{4}\)

\(B=\frac{1}{2}+\frac{-3}{2}-\frac{-3}{4}\)

\(B=\frac{2}{4}+\frac{-6}{4}-\frac{-3}{4}\)

\(\Rightarrow B=\frac{2+-6+3}{4}=\frac{-1}{4}\)

c, \(C=\frac{\left(\frac{1}{2}-0,75\right)\times\left(0,2-\frac{2}{5}\right)}{\frac{5}{9}-1\frac{1}{12}}\)

\(C=\frac{\left(\frac{1}{2}-\frac{3}{4}\right)\times\left(\frac{1}{5}-\frac{2}{5}\right)}{\frac{5}{9}-\frac{1\times12+1}{12}}\)

\(C=\frac{\left(\frac{2}{4}-\frac{3}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{5}{9}-\frac{13}{12}}\)

\(C=\frac{\left(\frac{-1}{4}\right)\times\left(\frac{-1}{5}\right)}{\frac{60}{108}-\frac{117}{108}}\)

\(C=\frac{\frac{1}{20}}{\frac{-19}{36}}=\frac{1}{20}\div\frac{-19}{36}=\frac{1}{20}\times\frac{36}{-19}\)

\(\Rightarrow C=\frac{36}{-380}=\frac{-9}{95}\)

d, \(D=\frac{\frac{2}{3}+\frac{2}{7}-\frac{1}{4}}{-1-\frac{3}{7}+\frac{3}{28}}\)

\(D=\frac{\frac{56}{84}+\frac{24}{84}-\frac{21}{84}}{\frac{-10}{7}+\frac{3}{28}}\)

\(D=\frac{\frac{59}{84}}{\frac{-40}{28}+\frac{2}{28}}=\frac{59}{84}\div\frac{-37}{28}=\frac{59}{84}\times\frac{28}{-37}\)

\(\Rightarrow D=\frac{1652}{-3108}=\frac{-59}{111}\)

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

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

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

9 tháng 8 2017

a) \(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\) \(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}=\frac{25}{33}\)

b) \(\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{10}{7}\right)=\left(1-\frac{1}{7}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{7}{7}\right).\left(1-\frac{8}{7}\right).\left(1-\frac{9}{7}\right).\) \(\left(1-\frac{10}{7}\right)\) = 0

9 tháng 8 2017

a)\(\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}+\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}+\frac{3}{293}}+\frac{\frac{7}{12}+\frac{5}{6}-1}{5-\frac{3}{4}+\frac{1}{3}}\)

\(=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}+\frac{1}{293}\right)}+\frac{\frac{7}{12}+\frac{10}{12}-\frac{12}{12}}{\frac{60}{12}-\frac{9}{12}+\frac{4}{12}}\)

\(=\frac{2}{3}+\frac{\frac{5}{12}}{\frac{55}{12}}\)

\(=\frac{2}{3}+\frac{1}{11}\)

\(=\frac{25}{33}\)

b)\(\left(1-\frac{1}{7}\right)\cdot\left(1-\frac{2}{7}\right)\cdot...\cdot\left(1-\frac{10}{7}\right)\)

Ta nhận thấy trong tích này có 1 thừa số là\(\left(1-\frac{7}{7}\right)=0\)nên tích trên sẽ bằng 0.

23 tháng 10 2016

a) \(A=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{\frac{5}{11}-\frac{5}{13}-\frac{5}{17}}+\frac{\frac{2}{3}-\frac{2}{9}-\frac{2}{27}+\frac{2}{81}}{\frac{7}{3}-\frac{7}{9}-\frac{7}{27}+\frac{7}{81}}\)

\(=\frac{\frac{1}{11}-\frac{1}{13}-\frac{1}{17}}{5\left(\frac{1}{11}-\frac{1}{13}-\frac{1}{17}\right)}+\frac{2\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}{7\left(\frac{1}{3}-\frac{1}{9}-\frac{1}{27}+\frac{1}{81}\right)}\)

\(=\frac{1}{5}+\frac{2}{7}\)

\(=\frac{7}{35}+\frac{10}{35}\)

\(=\frac{17}{35}\)

Vậy \(A=\frac{17}{35}\)

b) \(B=\frac{5^2}{11.16}+\frac{5^2}{16.21}+\frac{5^2}{21.26}+\frac{5^2}{26.31}+...+\frac{5^2}{56.61}\)

\(=5.\left(\frac{5}{11.16}+\frac{5}{16.21}+\frac{5}{21.26}+...+\frac{5}{56.61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+\frac{1}{21}-\frac{1}{26}+...+\frac{1}{56}-\frac{1}{61}\right)\)

\(=5.\left(\frac{1}{11}-\frac{1}{61}\right)\)

\(=5.\left(\frac{61}{671}-\frac{11}{671}\right)\)

\(=5.\frac{50}{671}\)

\(=\frac{250}{671}\)

Vậy \(B=\frac{250}{671}\)

26 tháng 3 2017

(2/5+2/7-2/11):(3/7-3/11+3/5) =2/5+2/7-2/11.7/3-11/3+5/3=2/1+2/1-2/1.1/3-1/3+1/3=2+1/3=7/3                                                                          Em đây mới học lớp 6 nên hay xem kĩ lại và tính bang máy tính  

16 tháng 4 2017

a, 2/3

b,1

nhớ nhé

13 tháng 11 2016

help me

25 tháng 4 2017

sao nhiều dữ vậy