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NH
15
29 tháng 8 2014
1/2+5/6+11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110= 1 - 1/2 + 1 - 1/6 + 1 - 1/12 .....+1 - 1/110= 10 - ( 1/2 + 1/6 + ...+ 1/110) = 10 - ( 1 - 1/ 2+ 1/2 - 1/ 3+ 1/3 - 1/4 ....+ 1/10 - 1/11)= 10 - (1 - 1/11)= 10 - 10/11 = 100/11
16 tháng 11 2020
a) \(\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{6}+1-\frac{1}{12}+...+1-\frac{1}{90}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\)
Từ 2 đến 9 có : ( 9 - 2 ) / 1 + 1 = 8 ( số hạng ) => có 8 số 1
\(\Rightarrow8-\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=8-\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=8-\frac{2}{5}=\frac{38}{5}\)
b) \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+...+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+...+1-\frac{1}{110}\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{110}\right)\)
\(=\left(1+1+...+1\right)-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{10\cdot11}\right)\)
Từ 1 đến 10 có : ( 10 - 1 ) / 1 + 1 = 10 ( số hạng ) => có 10 số 1
\(\Rightarrow10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{11}\right)\)
\(=10-\frac{10}{11}=\frac{100}{11}\)
NK
1
5 tháng 5 2018
11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110 =
256 phần 33
NK
1
5 tháng 5 2018
11/12+19/20+29/30+41/42+55/56+71/72+89/90+109/110
= \(\frac{256}{53}\)
F
2
16 tháng 1 2019
\(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}\)
\(=1-\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)
\(=1-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=1-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)\(=1-\left(\frac{1}{3}-\frac{1}{11}\right)=1-\frac{8}{33}=\frac{25}{33}\)
20 tháng 6 2016
5/6 + 11/12 + 19/20 + ... + 109/110
= (1 - 1/6) + (1 - 1/12) + (1 - 1/20) + ... + (1 - 1/110)
= (1 - 1/2×3) + (1 - 1/3×4) + (1 - 1/4×5) + ... + (1 - 1/10×11)
= (1 + 1 + 1 + ... + 1) - (1/2×3 + 1/3×4 + 1/4×5 + ... + 1/10×11)
( có 9 số 1)
= 1 x 9 - (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/10 - 1/11)
= 9 - (1/2 - 1/11)
= 9 - (11/22 - 2/22)
= 9 - 9/22
= 198/22 - 9/22
= 189/22
☆☆☆☆☆
DN
18 tháng 6 2015
\(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}+\frac{55}{56}+\frac{71}{72}+\frac{89}{90}+\frac{109}{110}\)
\(=\frac{12-1}{12}+\frac{20-1}{20}+\frac{30-1}{30}+\frac{42-1}{42}+\frac{56-1}{56}+\frac{72-1}{72}+\frac{90-1}{90}+\frac{110-1}{110}\)
\(=1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}+1-\frac{1}{56}+1-\frac{1}{72}+1-\frac{1}{90}+1-\frac{1}{110}\)
\(=8-\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=8-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=8-\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{256}{33}\)
18 tháng 6 2015
ĐẶT A = \(\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+...+\frac{109}{110}\)
8 -A = \(1-\frac{11}{12}+1-\frac{19}{20}+1-\frac{29}{30}+1-\frac{41}{42}+1-\frac{55}{56}+1-\frac{71}{72}+1-\frac{89}{90}+1-\frac{109}{110}\)
8 -A \(=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{10.11}\)
8 -A = \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
8 - A = \(\frac{1}{3}-\frac{1}{11}=\frac{8}{33}\)
=> A = \(8-\frac{8}{33}=\frac{256}{33}\)
Q
3 tháng 7 2016
A = \(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+......+\frac{109}{110}\)
A = \(1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}....+1-\frac{1}{110}\)
A = \(10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)
A = \(10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)
A = \(10-\left(1-\frac{1}{11}\right)\)
A = \(10-\frac{10}{11}\)
A = \(\frac{100}{11}\)