\(\frac{3}{2}+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+...............+\frac{91}{90}+\frac{111}{110}\)
K
Khách
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12 tháng 8 2019
\(A=\frac{3}{2}-\frac{5}{6}+\frac{13}{12}-\frac{19}{20}+\frac{31}{30}-\frac{41}{42}+\frac{57}{56}-\frac{71}{72}+\frac{91}{90}-\frac{109}{110}\)
\(\Rightarrow A=\left(1+\frac{1}{2}\right)-\left(1-\frac{1}{6}\right)+\cdot\cdot\cdot+\left(1+\frac{1}{90}\right)-\left(1-\frac{1}{110}\right)\)
\(\Rightarrow A=1+\frac{1}{2}-1+\frac{1}{6}+\cdot\cdot\cdot+1+\frac{1}{90}-1+\frac{1}{110}\)
\(\Rightarrow A=\left[\left(1-1\right)+\frac{1}{2}+\frac{1}{6}\right]+\cdot\cdot\cdot+\left[\left(1-1\right)+\frac{1}{90}+\frac{1}{110}\right]\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{6}+\cdot\cdot\cdot+\frac{1}{90}+\frac{1}{110}\)
\(\Rightarrow A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\cdot\cdot\cdot+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdot\cdot\cdot+\frac{1}{10}-\frac{1}{11}\)
\(\Rightarrow A=1-\frac{1}{11}\)
\(\Rightarrow A=\frac{10}{11}\)
D
1 tháng 5 2015
= \(\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{6}\right)+\left(1+\frac{1}{12}\right)+....+\left(1+\frac{1}{90}\right)\)
= \(\left(1+1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}\right)\)(9 số 1)
= 9 + \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{9.10}\right)\)
= \(9+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
= \(9+\left(1-\frac{1}{10}\right)=9+\frac{9}{10}=\frac{90}{10}+\frac{9}{10}=\frac{99}{10}\)
T
2 tháng 9 2020
\(\frac{5}{12}+\frac{5}{20}+\frac{5}{30}+...+\frac{5}{9900}=\frac{5}{3.4}+\frac{5}{4.5}+\frac{5}{5.6}+...+\frac{5}{99.100}\)
\(5\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(5\left(\frac{1}{3}-\frac{1}{100}\right)=\frac{97}{60}\)
11 tháng 7 2018
\(2x+\frac{7}{6}+\frac{13}{12}+\frac{21}{20}+\frac{31}{30}+\frac{43}{42}+\frac{57}{56}+\frac{73}{72}+\frac{91}{90}=10\)
=> \(2x+\frac{6+1}{6}+\frac{12+1}{12}+....+\frac{90+1}{90}=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+10=10\)
=> \(2x+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}=0\)
=>\(2x+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}=0\)
=>\(2x+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=0\)
=> \(2x-\frac{1}{10}=0\)
=>2x=\(\frac{1}{10}\)=> x=1/20
11 tháng 7 2018
mình có bị nhầm chỗ dấu suy ra thứ 3. đáng lẽ ra biểu thức đó cộng 8 chứ k phải cộng 10 do mình sơ ý nên bạn hãy sủa lại chỗ ấy
S
24 tháng 5 2017
Ta có:
\(A=\frac{3}{2}+\frac{13}{12}+\frac{31}{30}+\frac{57}{56}+\frac{91}{90}\)
\(=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)
\(=\left(1+1+1+1+1\right)+\left(\frac{1}{2}+\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+\frac{1}{90}\right)\)
\(=5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\)
\(B=\frac{5}{6}+\frac{19}{20}+\frac{41}{42}+\frac{71}{72}+\frac{109}{110}\)
\(=\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)
\(=\left(1+1+1+1+1\right)-\left(\frac{1}{6}+\frac{1}{20}+\frac{1}{42}+\frac{1}{72}+\frac{1}{110}\right)\)
\(=5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\)
=> A - B =\(\left[5+\left(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}\right)\right]-\left[5-\left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\right)\right]\)
= \(5+\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+\frac{1}{9.10}-5+\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{10.11}\)
= \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
= \(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
24 tháng 5 2017
\(A=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{12}\right)+\left(1+\frac{1}{30}\right)+\left(1+\frac{1}{56}\right)+\left(1+\frac{1}{90}\right)\)
\(B=\left(1-\frac{1}{6}\right)+\left(1-\frac{19}{20}\right)+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{72}\right)+\left(1-\frac{1}{110}\right)\)
Mk gợi ý đến đây thôi , mk bí rồi đợi mk nghĩ đã!
BM
10 tháng 4 2019
Đè thừa một số \(\frac{25}{156}\),mk ko lại đề bài nhé
\(A=1-\frac{2+3}{2\cdot3}+.....+\frac{11+12}{11\cdot12}-\frac{12+13}{12\cdot13}\)
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+\frac{1}{4}-...+\frac{1}{11}+\frac{1}{12}-\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{2}-\frac{1}{13}=\frac{11}{26}\)
Gọi tổng dãy số hạng trên là A
A = 1 + \(\frac{1}{2}\)+ 1 + \(\frac{1}{6}\)+ 1 + \(\frac{1}{12}\)+ ... + 1 + \(\frac{1}{90}\)+ 1 + \(\frac{1}{110}\)
Mà từ \(\frac{1}{2}\)đén \(\frac{1}{110}\) có 10 số
A = 1 x 10 + \(\frac{1}{2}\)+( \(\frac{1}{2}\)- \(\frac{1}{3}\)) + ( \(\frac{1}{3}\)-\(\frac{1}{4}\)) + (\(\frac{1}{4}\)-\(\frac{1}{5}\)) + ... + \(\frac{1}{11}\)
A = 10 + \(\frac{1}{2}\)+ \(\frac{1}{2}\)+ \(\frac{1}{11}\)= \(\frac{112}{11}\)