K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

27 tháng 4 2019

(1/57 + 1/5757 + 1/23 ) . ( 1/2 - 1/3 - 1/6 )

= ( 1/57 + 1/5757 + 1/23 ) . ( 3/6 - 2/6 - 1/6 )

= (1/57 + 1/5757 + 1/23 )  .  0

= 0

27 tháng 4 2019

\(\left(\frac{1}{57}+\frac{1}{5757}+\frac{1}{23}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{57}+\frac{1}{5757}+\frac{1}{23}\right).0\)

\(=0\)

chúc bạn học tốt

31 tháng 3 2019

a) \(\frac{53}{101}.\frac{-13}{97}+\frac{53}{101}.\frac{-84}{97}\)

\(=\frac{53}{101}\left(\frac{-13}{97}+\frac{-84}{97}\right)\)

\(=\frac{53}{101}.\frac{-97}{97}\)

\(=\frac{53}{101}.\left(-1\right)\)

\(=\frac{-53}{101}\)

b) \(\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{57}-\frac{1}{5757}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)

\(=\left(\frac{1}{57}-\frac{1}{5757}\right).0\)

\(=0\)

31 tháng 3 2019

c) \(\frac{3^2}{25}.\frac{75}{-21}.\frac{50}{35}\)

\(=\frac{3^2.75.50}{25.\left(-21\right).35}\)

\(=\frac{3.3.25.3.5.5.2}{25.3.\left(-7\right).5.7}\)

\(=\frac{3.3.5.2}{\left(-7\right).7}\)

\(=\frac{90}{-49}\)

d) \(\frac{25.48-25.18}{20.5^3}\)

\(=\frac{25\left(48-18\right)}{10.2.125}\)

\(=\frac{25.10.3}{10.2.25.5}\)

\(=\frac{3}{10}\)

a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)

b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)

c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)

d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)

a) \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\left(\frac{-3}{7}+\frac{3}{7}\right)+\left(\frac{15}{26}-\frac{2}{13}\right)\)

\(=\frac{15-4}{26}=\frac{11}{26}\)

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

\(=\frac{-11}{23}.2-\frac{1}{23}=\frac{-22-1}{23}=\frac{-23}{23}=-1\)

27 tháng 2 2018

Ta có : 

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.....\frac{99^2}{99.100}\)

\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1.2.2.3.3.4.....99.100}\)

\(=\)\(\frac{1^2.2^2.3^2.....99^2}{1^2.2^2.3^2.4^2.....99^2}.\frac{1}{100}\)

\(=\)\(\frac{1}{100}\)

30 tháng 5 2016

a. \(\frac{-3}{7}+\frac{15}{26}-\left(\frac{2}{13}-\frac{3}{7}\right)=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}=\frac{15}{13.2}-\frac{2}{13}=\frac{15}{13.2}-\frac{2.2}{13.2}=\frac{15-4}{26}=\frac{11}{26}\)

 

30 tháng 5 2016

C. \(\frac{-11}{23}.\frac{6}{7}+\frac{8}{7}.\frac{-11}{23}-\frac{1}{23}=\frac{1}{23}.\left(-11.\frac{6}{7}-11.\frac{8}{7}-1\right)=\frac{1}{23}.\left(-22-1\right)=\frac{1}{23}.\left(-23\right)=-1\)

15 tháng 6 2018

1)

\(2.\frac{3}{7}+\left(\frac{2}{9}-1\frac{3}{7}\right)-\frac{5}{3}:\frac{1}{9}\)

\(=\frac{6}{7}+\frac{2}{9}-\frac{10}{7}-\frac{5}{3}.9\)

\(=\left(\frac{6}{7}-\frac{10}{7}\right)+\frac{2}{9}-15\)

\(=\frac{-4}{7}+\frac{2}{9}-\frac{15}{1}\)

\(=\frac{-36}{63}+\frac{14}{63}-\frac{945}{63}\)

\(=\frac{-967}{63}\)

2)

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

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

\(=\frac{-11}{23}.2-\frac{1}{23}\)

\(=\frac{-22}{23}-\frac{1}{23}\)

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

\(=-1\)

3)

\(\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{24}\right)\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\left(\frac{4}{24}-\frac{3}{24}-\frac{1}{24}\right)\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).\frac{0}{24}\)

\(=\left(\frac{377}{-231}-\frac{123}{89}+\frac{34}{791}\right).0\)

\(=0\)

15 tháng 8 2019

\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)

\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)

\(A=\left[35-0\right]-5\frac{7}{32}\)

\(A=35-5\frac{7}{32}\)

\(A=\frac{953}{32}\)

\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)

\(B=71\frac{38}{45}-\frac{36377}{855}\)

\(B=\frac{1670}{57}\)

\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)

\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)

\(C=\frac{153}{14}:\frac{4}{5}\)

\(C=\frac{765}{56}\)

\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)

\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)

\(D=0-\frac{1}{4}\)

\(D=-\frac{1}{4}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)

\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)

\(\)\(E=\frac{22}{45}\)

CHUC BAN HOC TOT >.<