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Khách
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PV
2
S
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
https://olm.vn/hoi-dap/detail/63463423750.html
vào tham khảo đi mk nhác trình bày
PV
1
DK
15
NK
1
TS
14 tháng 11 2016
\(\frac{3}{12}+\frac{3}{20}+\frac{3}{30}+\frac{3}{42}+\frac{3}{56}+\frac{3}{72}+\frac{3}{90}+\frac{3}{110}\)
\(=\frac{3}{3\cdot4}+\frac{3}{4\cdot5}+\frac{3}{5\cdot6}+\frac{3}{6\cdot7}+\frac{3}{7\cdot8}+\frac{3}{8\cdot7}+\frac{3}{8\cdot9}+\frac{3}{9\cdot10}+\frac{3}{10\cdot11}\)
\(=3\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}+\frac{1}{9\cdot10}+\frac{1}{10\cdot11}\right)\)
\(=3\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{10}-\frac{1}{11}\right)\)
\(=3\cdot\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=3\cdot\left(\frac{11}{33}-\frac{3}{33}\right)\)
\(=3\cdot\frac{8}{33}\)
\(=\frac{24}{33}\)
\(=\frac{8}{11}\)
LH
2
LT
1
NC
8 tháng 6 2021
\(\frac{3}{2}-\frac{5}{6}+\frac{7}{12}-\frac{9}{20}+\frac{11}{30}-\frac{13}{42}+\frac{15}{56}-\frac{17}{72}+\frac{19}{90}\)
\(=\frac{1+2}{1.2}-\frac{2+3}{2.3}+\frac{3+4}{3.4}-\frac{4+5}{4.5}+\frac{5+6}{5.6}-\frac{6+7}{6.7}+\frac{7+8}{7.8}-\frac{8+9}{8.9}+\frac{9+10}{9.10}\)
\(=1+\frac{1}{2}-\frac{1}{2}+\frac{1}{3}+\frac{1}{4}-...-\frac{1}{8}-\frac{1}{9}+\frac{1}{9}+\frac{1}{10}\)
\(=\frac{11}{10}\)
TL
3
16 tháng 8 2018
a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)
\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)
\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)
b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)
\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)
c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)
\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)
\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)
\(\Rightarrow x=9\)
DD
Đoàn Đức Hà
Giáo viên
10 tháng 1 2021
b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{15}\)
C = 2,625
\(\frac{3}{2}+\frac{3}{6}+\frac{3}{12}+\frac{3}{20}+\frac{3}{30}+\frac{3}{42}+\frac{3}{56}\)
\(\frac{3}{1\cdot2}+\frac{3}{2\cdot3}+\frac{3}{3\cdot4}+\frac{3}{4\cdot5}+\frac{3}{5\cdot6}+\frac{3}{6\cdot7}+\frac{3}{7\cdot8}\)
\(3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(3\cdot\left(1-\frac{1}{8}\right)\)
\(3\cdot\frac{7}{8}=\frac{21}{8}\)