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NT
3
27 tháng 6 2018
\(\frac{3}{7}\times\frac{9}{25}-\frac{1}{14}\times\frac{1}{13}\)
\(=\frac{27}{175}-\frac{1}{182}\)
\(=\frac{702}{4550}-\frac{25}{4550}\)
\(=\frac{677}{4550}\)
27 tháng 6 2018
\(\frac{3}{7}.\frac{9}{25}-\frac{1}{14}.\frac{1}{13}\)
\(=\frac{27}{175}-\frac{1}{182}\)
\(=\frac{702}{4550}-\frac{25}{4550}\)
\(=\frac{677}{4550}\)
3 tháng 8 2015
a, \(\frac{-3}{7}+\frac{5}{13}-\frac{4}{7}+\frac{8}{13}\)
\(=\frac{-3}{7}-\frac{4}{7}+\frac{5}{13}+\frac{8}{13}\)
\(=-\frac{7}{7}+\frac{13}{13}=-1+1=0\)
b, \(\frac{-5}{14}-\frac{2}{-14}+\frac{1}{8}+\frac{1}{8}\)
\(=\frac{-5}{14}+\frac{2}{14}+\frac{1}{8}+\frac{1}{8}\)
\(=-\frac{3}{14}+\frac{1}{4}=\frac{1}{28}\)
c,\(-\frac{5}{13}-\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)\)
\(=-\frac{5}{13}-\frac{3}{13}-\frac{3}{5}+\frac{4}{10}\)
\(=-\frac{8}{13}-\frac{3}{5}+\frac{4}{10}=-\frac{79}{65}+\frac{4}{10}=-\frac{53}{65}\)
d, \(\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{4}{6}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{9}{7}+\frac{2}{3}-\frac{12}{7}-\frac{1}{2}\right)+\frac{5}{9}\right]\)
\(=\left[\left(\frac{1}{8}-\frac{1}{2}-\frac{9}{7}-\frac{12}{7}+\frac{2}{3}\right)+\frac{5}{9}\right]\)
\(=-\frac{65}{24}+\frac{5}{9}=-2\frac{11}{72}\)
7 tháng 10 2024
a)-3/7+5/13-4/7+8/13
=-3/7-4/7+5/13+8/13
=-7/7+13/13
=-1+1
=0
PT
1
HH
26 tháng 6 2018
Giải:
a) \(\dfrac{1}{3}.\left(-\dfrac{4}{5}\right)+\dfrac{1}{3}.\left(-\dfrac{6}{5}\right)\)
\(=\dfrac{1}{3}\left[-\dfrac{4}{5}+\left(-\dfrac{6}{5}\right)\right]\)
\(=\dfrac{1}{3}\left[-\dfrac{4}{5}-\dfrac{6}{5}\right]\)
\(=\dfrac{1}{3}\left(-\dfrac{10}{5}\right)\)
\(=\dfrac{-10}{15}=-\dfrac{2}{3}\)
Vậy ...
b) \(\dfrac{3}{7}.\dfrac{9}{25}-\dfrac{1}{14}.\dfrac{1}{13}\)
\(=\dfrac{1}{7}.\dfrac{27}{25}-\dfrac{1}{7}.\dfrac{1}{26}\)
\(=\dfrac{1}{7}\left(\dfrac{27}{25}-\dfrac{1}{26}\right)\)
\(=\dfrac{1}{7}.\dfrac{677}{650}\)
\(=\dfrac{677}{4550}\)
Vậy ...
HG
3
18 tháng 8 2018
hot girl á cậu là hốt rác thì có đòi làm hot girl nhưng cậu chỉ làm được hốt rác thôi
G
13 tháng 9 2020
a)\(\frac{7}{12}.\frac{6}{11}+\frac{7}{12}.\frac{5}{11}-2\frac{7}{12}\)
\(=\frac{7}{12}.\left(\frac{6}{11}+\frac{5}{11}\right)-\frac{31}{12}\)
\(=\frac{7}{12}-\frac{31}{12}\)
\(=-2\)
b)\(\frac{-5}{9}.\frac{-6}{13}+\frac{5}{-9}.\frac{-5}{13}-\frac{5}{9}\)
\(=\frac{5}{9}.\left(\frac{6}{13}+\frac{5}{13}-1\right)\)
\(=\frac{5}{9}.\left(\frac{11}{13}-\frac{13}{13}\right)\)
\(=\frac{5}{9}.\frac{-2}{13}\)
\(=-\frac{10}{117}\)
c)\(0,8.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-1\frac{2}{5}\)
\(=\frac{4}{5}.\frac{-15}{14}-\frac{4}{5}.\frac{13}{14}-\frac{7}{5}\)
\(=\frac{4}{5}.\left(-\frac{15}{14}-\frac{13}{14}\right)-\frac{7}{5}\)
\(=\frac{4}{5}.\left(-2\right)-\frac{7}{5}\)
\(=\frac{-8}{5}-\frac{7}{5}\)
\(=-3\)
d)\(-75\%.\frac{6}{7}+5\%.\frac{6}{7}+\frac{7}{10}.1\frac{1}{7}\)
\(=\frac{-15}{20}.\frac{6}{7}+\frac{1}{20}.\frac{6}{7}+\frac{7}{10}.\frac{8}{7}\)
\(=\frac{6}{7}.\left(\frac{-15}{20}+\frac{1}{20}\right)+\frac{4}{5}\)
\(=\frac{6}{7}.\frac{-7}{10}+\frac{4}{5}\)
\(=-\frac{3}{5}+\frac{4}{5}\)
\(=\frac{1}{5}\)
Linz
13 tháng 9 2020
\(B=\frac{1}{5}-\frac{3}{7}+\frac{5}{9}-\frac{2}{11}+\frac{7}{13}-\frac{9}{16}-\frac{7}{13}+\frac{2}{11}-\frac{5}{9}+\frac{3}{7}-\frac{1}{5}\)
\(=\frac{1}{5}-\frac{1}{5}-\frac{3}{7}+\frac{3}{7}+\frac{5}{9}-\frac{5}{9}-\frac{2}{11}+\frac{2}{11}+\frac{7}{13}-\frac{7}{13}-\frac{9}{16}\)
\(=0+0+0+0+0-\frac{9}{16}\)
\(=\frac{-9}{16}\)