\(C=-\dfrac{3}{11.14}-\dfrac{3}{14.17}-\dfrac{3}{17.20}-...-\dfrac{3}{98.101}...">
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7 tháng 9 2022

\(C=-\dfrac{3}{11.14}-\dfrac{3}{14.17}-\dfrac{3}{17.20}-...-\dfrac{3}{98.101}\)

\(\Rightarrow C=-\left(\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+...+\dfrac{3}{98.101}\right)\)

\(\Rightarrow C=-\left(\dfrac{14-11}{11.14}+\dfrac{17-14}{14.17}+...+\dfrac{101-98}{98.101}\right)\)

\(\Rightarrow C=-\left(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{98}-\dfrac{1}{101}\right)\)

\(\Rightarrow C=-\left(\dfrac{1}{11}-\dfrac{1}{101}\right)=\dfrac{-90}{1111}\)

7 tháng 9 2022

Cảm ơn bạn! Đúng lúc bạn đang trả lời thì mình lại biết cách làm rồi. Cũng giống như cách làm của bạn.

 

3 tháng 3 2017

\(\frac{15}{11.14}+\frac{15}{14.17}+\frac{15}{17.20}+...+\frac{15}{72.75}\)

\(=5\left(\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}+...+\frac{3}{72.75}\right)\)

\(=5\left(\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17}+\frac{1}{17}-\frac{1}{20}+...+\frac{1}{72}-\frac{1}{75}\right)\)\(=5\left(\frac{1}{11}-\frac{1}{75}\right)\)

\(=\frac{64}{165}\)

3 tháng 3 2017

pài này gần giống pài troq v15

17 tháng 8 2017

\(\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{68.71}\)

\(=5\left(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{68}-\dfrac{1}{71}\right)\)

\(=5\left(\dfrac{1}{11}-\dfrac{1}{71}\right)\)

\(=5.\dfrac{60}{781}\)

\(=\dfrac{300}{781}\)

12 tháng 3 2017

theo bài ra ta có:

\(E=\dfrac{15}{11.14}+\dfrac{15}{14.17}+\dfrac{15}{17.20}+...+\dfrac{15}{74.77}\\ \Rightarrow\dfrac{1}{5}E=\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+...+\dfrac{3}{74.77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{20}+...+\dfrac{1}{74}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{1}{11}-\dfrac{1}{77}\\ \dfrac{1}{5}E=\dfrac{7}{77}-\dfrac{1}{77}=\dfrac{6}{77}\\ \Rightarrow E=\dfrac{6}{77}.5\\ E=\dfrac{30}{77}\)

5 .\((\)\(\dfrac{3}{11.14}+\dfrac{3}{14.17}+...+\dfrac{3}{74.77}\))

= 5. (\(\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}+...+\dfrac{1}{74}-\dfrac{1}{77}\))

= 5.(\(\dfrac{1}{11}-\dfrac{1}{77}\))

= 5. \(\dfrac{6}{77}\)

= \(\dfrac{30}{77}\)

2 tháng 8 2018

a, 1/3-3/4+3/5+1/4-2/9-1/36+1/15

=(1/3+3/5+1/15)-(3/4-1/4+2/9+1/36)

=1 - 3/4

=1/4

b, 3-1/4+2/3-5-1/3+6/5-6+7/4-3/2

=(3-5-6)-(1/4-7/4)+(2/3-1/3)+(6/5-3/2)

=-8 +3/2 +1/3 -3/10

=-97/15

7 tháng 10 2017

a) \(\dfrac{-5}{9}.\dfrac{3}{11}+\dfrac{-13}{18}.\dfrac{3}{11}\)

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

\(=\dfrac{3}{11}.\left(-2\right)\)

\(=\dfrac{-6}{11}\)

b) \(\dfrac{11}{2}.2\dfrac{1}{3}-1\dfrac{1}{5}.1\dfrac{1}{2}\)

\(=\dfrac{11}{3}.\dfrac{7}{3}-\dfrac{6}{5}.\dfrac{3}{2}\)

\(=\dfrac{77}{9}-\dfrac{9}{5}\)

\(=\dfrac{385}{45}-\dfrac{81}{45}\)

\(=\dfrac{304}{45}\)

c) \(1\dfrac{1}{9}.\dfrac{2}{145}-4\dfrac{1}{3}-\dfrac{2}{145}+\dfrac{2}{145}\)

\(=\dfrac{10}{9}.\dfrac{2}{145}-\dfrac{8}{3}\)

\(=\dfrac{4}{261}-\dfrac{8}{3}\)

\(=\dfrac{4}{261}-\dfrac{696}{261}\)

\(=-\dfrac{692}{261}\)

d) \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)

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

\(=0+0+0+4-1-1-1\)

\(=4-3\)

\(=1\)

31 tháng 8 2017

a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)

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

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

= \(\dfrac{3}{11}\)

b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)

= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)

= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)

= \(\dfrac{3}{5}\)

c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)

= \(-1,5.\left(\dfrac{4}{3}-2\right)-\dfrac{2}{3}\)

= \(-1,5.\left(\dfrac{-2}{3}\right)-\dfrac{2}{3}\)

= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)

Bài 1:

a: \(=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{4}{3}-1+\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)

b: \(=\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{1}{9}-1-\dfrac{2}{5}+\dfrac{5}{4}=2-1+\dfrac{1}{9}=\dfrac{10}{9}\)

c: \(=\left(\dfrac{-3}{2}\cdot\dfrac{4}{3}\right)\cdot\dfrac{-9}{2}-\dfrac{1}{2}=9-\dfrac{1}{2}=8.5\)

20 tháng 6 2018

a/Ta có: \(\dfrac{4}{3}-\left[\left(\dfrac{-11}{6}\right)-\left(\dfrac{2}{9}+\dfrac{5}{3}\right)\right]\)

\(=\) \(\dfrac{4}{3}-\left[\dfrac{-11}{6}-\dfrac{2}{9}-\dfrac{5}{3}\right]\)

\(=\) \(\dfrac{4}{3}+\dfrac{11}{6}+\dfrac{2}{9}+\dfrac{5}{3}\)

\(=\) \(\dfrac{24}{18}+\dfrac{33}{18}+\dfrac{4}{18}+\dfrac{30}{18}\)

\(=\) \(\dfrac{91}{18}\)

b/Ta có: \(\left(8-\dfrac{9}{4}+\dfrac{2}{7}\right)-\left(-6-\dfrac{3}{7}+\dfrac{5}{4}\right)-\left(3+\dfrac{2}{4}-\dfrac{9}{7}\right)\)

\(=\) \(8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)

\(=\) \(8+6-3-\dfrac{9}{4}-\dfrac{5}{4}-\dfrac{2}{4}+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}\)

\(=\) \(11-\dfrac{2}{4}+\dfrac{14}{7}\)

\(=\) \(11-\dfrac{1}{2}+2\)

\(=\) \(9-\dfrac{1}{2}\)

\(=\) \(\dfrac{17}{2}\)

Chúc bn học tốt!!!okokok

22 tháng 6 2018

phần b thầy mk chữa là = 9 đó

những dù sao cx thank youvui

10 tháng 10 2018

A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)

12 tháng 10 2022

\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)

\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)

=1-82/84

=2/84=1/42

\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)

\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)

\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)

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