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14 tháng 11 2016

\(\frac{3}{12}+\frac{3}{20}+\frac{3}{30}+\frac{3}{42}+\frac{3}{56}+\frac{3}{72}+\frac{3}{90}+\frac{3}{110}\)

\(=\frac{3}{3\cdot4}+\frac{3}{4\cdot5}+\frac{3}{5\cdot6}+\frac{3}{6\cdot7}+\frac{3}{7\cdot8}+\frac{3}{8\cdot7}+\frac{3}{8\cdot9}+\frac{3}{9\cdot10}+\frac{3}{10\cdot11}\)

\(=3\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\right)\)

\(=3\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=3\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=3\cdot\left(\frac{11}{33}-\frac{3}{33}\right)\)

\(=3\cdot\frac{8}{33}\)

\(=\frac{24}{33}\)

\(=\frac{8}{11}\)

20 tháng 2 2019

\(\frac{3}{12}+\frac{3}{20}+\frac{3}{30}+...+\frac{3}{110}\)

\(=3\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{110}\right)\)

\(=3\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{10\cdot11}\right)\)

\(=3\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{10}-\frac{1}{11}\right)\)

\(=3\left(\frac{1}{3}-\frac{1}{11}\right)\)

\(=3\cdot\frac{8}{33}=\frac{8}{11}\)

20 tháng 2 2019

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vào tham khảo đi mk nhác trình bày

20 tháng 2 2019

= 3.(1/12+1/20+1/30+1/42+1/56+1/72+1/90+1/110)

= 3.8/33

= 8/11

25 tháng 10 2023

rút gọn thành

1/2 x 2/3 x 3/4 x 4/5 x 5/6 x 6/7 x 7/8 x 8/9

1x2x3x4x5x6x7x8

__________________

2x3x4x5x6x7x8x9

gạch các số giống nhau ở tử và mẫu và bằng 1/9

 

 

19 tháng 6 2018

Giải:

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-5}{11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-5}{11}+\dfrac{5}{13}=x\)

\(\Leftrightarrow x=\dfrac{-10}{143}\)

Vậy ...

19 tháng 6 2018

Giải đẳng cấp nhỉ @Hắc Hường

14 tháng 7 2015

A - B = \(\left(1+\frac{1}{2}+1+\frac{1}{12}+1+\frac{1}{30}+1+\frac{1}{56}+1+\frac{1}{90}\right)-\left(1-\frac{1}{6}+1-\frac{1}{20}+1-\frac{1}{42}+1-\frac{1}{72}+1-\frac{1}{110}\right)\)\(\left(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)-\left(5-\frac{1}{6}-\frac{1}{20}-\frac{1}{42}-\frac{1}{72}-\frac{1}{110}\right)\)\

\(5+\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}-5+\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\)

\(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=1-\frac{1}{11}=\frac{10}{11}\)

 

15 tháng 10 2018

\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}+\frac{109}{110}\)

\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}+1-\frac{1}{110}\)

\(=10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\right)\)

\(=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\right)\)

\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)

\(=10-\left(1-\frac{1}{10}\right)\)

\(=\frac{91}{10}\)

24 tháng 7 2018

\(-\frac{3}{20}+\frac{-3}{30}+\frac{-3}{42}+\frac{-3}{56}+\frac{-3}{72}+\frac{-3}{90}\)

\(=-3\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=-3\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{10}\right)=-\frac{9}{20}\)

24 tháng 7 2018

\(\frac{-3}{20}+\frac{-3}{30}+\frac{-3}{42}+\frac{-3}{56}+\frac{-3}{72}+\frac{-3}{90}\)

\(=-3\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\right)\)

\(=-3\left(\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{9\cdot10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}\right)\)

\(=-3\left(\frac{1}{4}-\frac{1}{10}\right)\)

\(=-3\cdot\frac{3}{20}\)

\(=\frac{-9}{20}\)

18 tháng 7 2017

\(\frac{1}{3}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}-\frac{1}{72}-\frac{1}{90}-\frac{1}{110}=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\frac{1}{3}+\frac{1}{11}=x-\frac{5}{13}\)

\(\frac{1}{11}=x-\frac{5}{13}\)

\(x=\frac{1}{11}+\frac{5}{13}\)

\(x=\frac{68}{143}\)