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19 tháng 4 2023

`3/13 . 15/11 + 3/13 . 7/11 -3/13`

`= 3/13 . 15/11 + 3/13 . 7/11 -3/13 .1`

`= 3/13 . (15/11 + 7/11 -11/11)`

`= 3/13 . 11/11`

`= 3/13 . 1`

`=3/13`

20 tháng 4 2023

M là Minh:)))

12 tháng 7 2021

a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

A=0+0+0+...+0+\(\dfrac{13}{15}\)

A=\(\dfrac{13}{15}\)

b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)

12 tháng 9 2023

\(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}+\dfrac{13}{15}+\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
\(=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\left(\dfrac{11}{13}-\dfrac{11}{13}\right)+\dfrac{13}{15}\)
\(=\dfrac{13}{15}\)

12 tháng 9 2023


=(13−13)−(35−35)+(57−57)−(79−79)+(911−911)−(1113−1113)+1315
=1315

 

Đề bài yêu cầu gì?

27 tháng 2 2022

\(a,15\dfrac{3}{13}-\left(3\dfrac{4}{7}+8\dfrac{3}{13}\right)=15\dfrac{3}{13}-3\dfrac{4}{7}-8\dfrac{3}{13}=\left(15\dfrac{3}{13}-8\dfrac{3}{13}\right)-\dfrac{25}{7}=7-\dfrac{25}{7}=\dfrac{49}{7}-\dfrac{25}{7}=\dfrac{24}{7}\)

\(b,\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}=\left(7\dfrac{4}{9}-3\dfrac{4}{9}\right)+4\dfrac{4}{9}=4+\dfrac{40}{9}=\dfrac{36}{9}+\dfrac{40}{9}=\dfrac{76}{9}\)

\(c,\dfrac{-7}{9}.\dfrac{4}{11}+\dfrac{-7}{9}.\dfrac{7}{11}+5\dfrac{7}{9}=\dfrac{-7}{9}\left(\dfrac{4}{11}+\dfrac{7}{11}\right)+\dfrac{52}{9}=\dfrac{-7}{9}.1+\dfrac{52}{9}=\dfrac{-7}{9}+\dfrac{52}{9}=\dfrac{45}{9}=5\)

\(d,50\%.1\dfrac{1}{3}.10.\dfrac{7}{35}.0,75=\dfrac{1}{2}.\dfrac{4}{3}.10.\dfrac{1}{5}.\dfrac{3}{4}=\left(\dfrac{1}{2}.\dfrac{1}{5}.10\right).\left(\dfrac{4}{3}.\dfrac{3}{4}\right)=1.1=1\)

\(e,\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{40.43}=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{40}-\dfrac{1}{43}=1-\dfrac{1}{43}=\dfrac{42}{43}\)

27 tháng 2 2022

Viết lại phần d) đc 0 ạ=((

26 tháng 6 2017

\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)

15 tháng 6 2018

\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}+\dfrac{13}{11}}\)

= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}{13.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{11}\right)}=\dfrac{3}{13}\)

24 tháng 6 2017

\(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{11}}{\dfrac{13}{4}-\dfrac{13}{5}+\dfrac{13}{7}-\dfrac{13}{11}}\\ =\dfrac{\dfrac{1311}{1540}}{\dfrac{2041}{1540}}\\ =\dfrac{1311}{2041}\)

24 tháng 6 2017

bn ơi có cách tính nào nhanh hơn ko bnbucminh

24 tháng 8 2018

Dấu "=" chính giữa mình ko biết là dấu + hay - nên mình sẽ làm cả hai TH nhé

TH1: \(P=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}+\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)

\(=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}+\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}+\dfrac{3}{11}=\dfrac{6}{11}\)

Tương tự TH2:

\(P=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}-\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3}{11}-\dfrac{3}{11}=0\)