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C=7/10x11+7/11x12+7/12x13+.................+7/69x70
C=1x7/10x11+1x7/11x12+...........+1x7/69x70
C=7(1/10x11+1/11x12+1/12x13+....+1/69x70)
C=7(1/10-1/11+1/11-1/12+1/12-1/13+.......+1/69-1/70)
C=7(1/10-1/70)
C=7(7/70-1/70)
C=7x6/70
C=3/5
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(=7.\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(=7.\frac{3}{35}=\frac{3}{5}\)
\(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=1\left(\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\right)\)
\(A=7\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\left(\frac{1}{10}-\frac{1}{70}\right)=7.\frac{3}{35}=\frac{3}{5}\)
C=7/10x11+7/11x12+7/12x13+.................+7/69x70
C=1x7/10x11+1x7/11x12+...........+1x7/69x70
C=7(1/10x11+1/11x12+1/12x13+....+1/69x70)
C=7(1/10-1/11+1/11-1/12+1/12-1/13+.......+1/69-1/70)
C=7(1/10-1/70)
C=7(7/70-1/70)
C=7x6/70
C=3/5
a) \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(3F=\frac{3}{3.6}+\frac{3}{6.9}+\frac{3}{9.12}+...+\frac{3}{30.33}\)
\(3F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{30}-\frac{1}{33}\)
\(3F=\frac{1}{3}-\frac{1}{33}\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\frac{1}{3}-\frac{1}{3}.\frac{1}{33}=\frac{1}{9}-\frac{1}{99}=\frac{11}{99}-\frac{1}{99}=\frac{10}{99}\)
b) \(A=\frac{7}{10.11}+\frac{7}{11.12}+\frac{7}{12.13}+...+\frac{7}{69.70}\)
\(A=7.\left(\frac{1}{10.11}+\frac{1}{11.12}+\frac{1}{12.13}+...+\frac{1}{69.70}\right)\)
\(A=7.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7.\left(\frac{1}{10}-\frac{1}{70}\right)=7.\left(\frac{7}{70}-\frac{1}{70}\right)=7.\frac{6}{70}\)
\(A=\frac{7.6}{70}=\frac{1.6}{10}=\frac{1.3}{5}=\frac{3}{5}\)
a, \(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3}\cdot\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{30\cdot33}\right)\)
\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}\cdot\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}-\frac{10}{33}\)
\(F=\frac{10}{99}\)
a: =(2/7-2/7)(-4/7-5/9)=0
b:
Sửa đề: 9/13*(-12/17)+9/13*29/27
=9/13(-12/17+29/17)
=9/13*17/17=9/13
c: \(=\dfrac{1}{7}\left(4+\dfrac{6}{7}+\dfrac{8}{7}\right)=\dfrac{1}{7}\cdot6=\dfrac{6}{7}\)
d: =7/10(5/7+9/7+8/7+13/7)
=5*7/10=7/2
a) Ta có: \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\)
\(=\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\)
\(=\dfrac{15}{26}-\dfrac{4}{26}\)
\(=\dfrac{11}{26}\)
b) Ta có: \(2\cdot\dfrac{3}{7}+\left(\dfrac{2}{9}-1\dfrac{3}{7}\right)-\dfrac{5}{3}:\dfrac{1}{9}\)
\(=\dfrac{6}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9\)
\(=\dfrac{-4}{7}+\dfrac{2}{9}-15\)
\(=\dfrac{-36}{63}+\dfrac{14}{63}-\dfrac{945}{63}\)
\(=\dfrac{-967}{63}\)
c) Ta có: \(\dfrac{-11}{23}\cdot\dfrac{6}{7}+\dfrac{8}{7}\cdot\dfrac{-11}{23}-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}\)
\(=\dfrac{-11}{23}\cdot2-\dfrac{1}{23}\)
\(=-1\)
d) Ta có: \(\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\left(\dfrac{4}{24}-\dfrac{3}{24}-\dfrac{1}{24}\right)\)
\(=\left(\dfrac{-377}{231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot0\)
=0
a)\(\dfrac{7}{-25}+-\dfrac{8}{25}=-\dfrac{15}{25}=-\dfrac{3}{5}\)
b)\(\dfrac{7}{21}-\dfrac{9}{-36}\)
\(=\dfrac{1}{3}+\dfrac{1}{4}\)
\(=\dfrac{4}{12}+\dfrac{3}{12}=\dfrac{7}{12}\)
c)\(-\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{1}{4}+\dfrac{5}{7}\)
\(=\left(-\dfrac{3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\)
\(=-\dfrac{1}{2}+1\)
\(=\dfrac{2}{2}-\dfrac{1}{2}=\dfrac{1}{2}\)
\(a,\dfrac{7}{-25}+\dfrac{-8}{25}=\dfrac{-7}{25}+\dfrac{-8}{25}=\dfrac{-15}{25}=\dfrac{-3}{5}\\ b,\dfrac{7}{21}-\dfrac{9}{-36}=\dfrac{7}{21}+\dfrac{9}{36}=\dfrac{7}{12}\\ c,\dfrac{-3}{4}+\dfrac{2}{7}+\dfrac{1}{4}+\dfrac{5}{7}\\ =\left(\dfrac{-3}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{7}+\dfrac{5}{7}\right)\\ =-\dfrac{1}{2}+1\\ =\dfrac{1}{2}\)
A = 6/19 . -7/11 + 6/19 . -4/11 + -13/19
A = 6/19 . [-7/11 + (-4/11)] + (-13/19)
A = 6/19 . -11/11 + (-13/19)
A = 6/19 . (-1) + (-13/19)
A = -6/19 + (-13/19)
A = -19/19
A = -1
Vậy A = -1
\(A=\dfrac{6}{19}.\dfrac{-7}{11}+\dfrac{6}{19}.\dfrac{-4}{11}+\dfrac{-13}{19}\)
\(A=\dfrac{6}{19}.\left(\dfrac{-7}{11}+\dfrac{-4}{11}\right)+\dfrac{-13}{19}\)
\(A=\dfrac{6}{19}.-1+\dfrac{-13}{19}=\dfrac{-6}{19}+\dfrac{-13}{19}=\dfrac{-19}{19}=-1\)
B=57.−413+57.73−513.37
B=57.−413+57.73− \(\dfrac{5}{7}.\dfrac{3}{13}\)3
7
\(B=\dfrac{5}{7}.\left(\dfrac{-4}{13}+\dfrac{7}{3}-\dfrac{3}{13}\right)\)
\(B=\dfrac{5}{7}.\dfrac{70}{39}=\dfrac{350}{273}\)
Câu B ko rõ lắm
\(B=\dfrac{5}{7}.\dfrac{-4}{13}+\dfrac{5}{7}.\dfrac{7}{3}-\dfrac{5}{13}.\dfrac{3}{7}\)
\(B=\dfrac{5}{7}.\dfrac{-4}{13}+\dfrac{5}{7}.\dfrac{7}{3}-\dfrac{5}{7}.\dfrac{3}{13}\)
Rồi tiếp 2 câu rõ
`A=7/(10*11)+7/(11*12)+7/(12*13)+...+7/(69*70)`
`1/7A=1/(10*11)+1/(11*12)+1/(12*13)+..+1/(69*70)`
`1/7A=1/10-1/11+1/11-1/12+1/12-1/13+...+1/69-1/70`
`1/7A=1/10-1/70`
`1/7A=7/70-1/70`
`1/7A=6/70`
`A=3/5`
\(A=\dfrac{7}{10.11}+\dfrac{7}{11.12}+\dfrac{7}{12.13}+...+\dfrac{7}{69.70}\)
\(A=7.\left(\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+...+\dfrac{1}{69.70}\right)\)
\(A=7\left(\dfrac{11-10}{10.11}+\dfrac{12-11}{11.12}+\dfrac{13-12}{12.13}+...+\dfrac{70-69}{69.70}\right)\)
\(A=7.\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(A=7.\left(\dfrac{1}{10}-\dfrac{1}{70}\right)\)
\(A=7.\dfrac{3}{35}=\dfrac{3}{5}\)