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HH
1
22 tháng 7 2015
\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{15}\right)=\frac{1}{2}\cdot\frac{4}{15}=\frac{2}{15}\)
28 tháng 4 2021
Ta có:A = 1/1.2 + 1/2.3 + 1/3.4 + ...+ 1/2020.2021
A=1-1/2+1/2-1/3+1/3-1/4+...+1/2020-1/2021
A=1-1/2021
28 tháng 4 2021
Ta có: B = 1/6 + 1/12 + 1/20 + ... + 1/240
B=1/2.3+1/3.4+1/4.5+....+1/15.16
B=1/2-1/3+1/3-1/4+1/4-1/5+....+1/15-1/16
B=1/2-1/16
phần C bn có đánh nhầm đề bài ko
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
4 tháng 9 2015
A = 1/2 + 1/3 + 1/6 + 1/12 + 1/15 + 1/20 + 1/30 + 1/35 + 1/42 + 1/56 + 1/63 + 1/72 + 1/99
= ( 1/2 + 1/6 + 1/12 + 1/20 + 1/30 + 1/42 + 1/56 + 1/72 ) + ( 1/3 + 1/15 + 1/35 + 1/63 + 1/99 )
= ( 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + ... + 1/8.9 ) + ( 1/1.3 + 1/3.5 + ... + 1/9.11 )
= ( 1 - 1/2 + 1/2 - 1/3 + ... + 1/8 - 1/9 ) + 1/2 . ( 2/1.3 + 2/3.5 + ... + 2/9.11 )
= ( 1 - 1/9 ) + 1/2 . ( 1 - 1/3 + 1/3 - 1.5 + ... + 1/9 - 1/11 )
= 8/9 + 1/2 . ( 1 - 1/11 )
= 8/9 + 1/2 . 10/11
= 8/9 + 5/11
= 133/99
20 tháng 7 2018
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
TN
20 tháng 7 2018
\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{10}{11}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\)
\(=1-\frac{1}{7}\)
\(=\frac{6}{7}\)
30 tháng 6 2020
\(B=\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}\)
\(=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+\frac{1}{11\cdot13}\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{13}\right)=\frac{5}{39}\)
25 tháng 11 2015
1-1/2+1/2-1/3+1/3+1/4-1/4+1/5-1/5+1/6-1/6+1/7-1/7+1/8-1/8+1/9-1/9+1/10-(1-1/3+1/3-3/5+3/5-4/7+5/9-5/9+6/11-6/11-7/13)=1+1/10-1+7/13=83/130
HC
2
25 tháng 11 2015
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}-\frac{1}{1.3}-\frac{1}{3.5}-\frac{1}{5.7}-...-\frac{1}{11.13}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{13}\right)\)
\(=1-\frac{1}{10}-\frac{1}{2}.\left(1-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{57}{130}\)
25 tháng 11 2015
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{90}-\frac{1}{3}-\frac{1}{15}-.....-\frac{1}{143}\)
\(=\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)-\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{143}\right)\)
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+.....+\frac{1}{9.10}\right)-\left(\frac{1}{1.3}+\frac{1}{3.5}+.....+\frac{1}{11.13}\right)\)
\(=\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-.....-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}\right)\)\(=\left(\frac{1}{1}-\frac{1}{10}\right)-\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{13}\right)=\frac{9}{10}-\frac{6}{13}=\frac{117}{130}-\frac{78}{130}=\frac{39}{130}=\frac{3}{10}\)
TH
0
\(a)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{22}{132}+\frac{11}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{33}{132}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{34}{132}+\frac{1}{20}\)
\(=\frac{17}{66}+\frac{1}{20}\)
\(=\frac{203}{660}\)
\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{132}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\left(\frac{1}{2}-\frac{1}{5}\right)+\frac{1}{132}\)
\(=\frac{3}{10}+\frac{1}{132}\)
\(=\frac{198}{660}+\frac{5}{660}\)
\(=\frac{203}{660}\)