LB
1
K
Khách
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T
3
5 tháng 8 2016
Ta có:
\(A=\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(A=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{87.90}\right)\)
\(A=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(A=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(A=2.\frac{1}{18}=\frac{1}{9}\)
N
0
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
29 tháng 4 2016
a/ \(=\frac{21}{23}+\frac{125}{143}-\frac{101.21}{101.23}-\frac{1001.125}{1001.143}=0\)
b/ \(=\frac{4}{20}+\frac{8}{21}+\frac{2}{5}-\frac{3}{5}+\frac{2}{21}-\frac{10}{21}+\frac{3}{20}=\frac{7}{20}-\frac{1}{5}=\frac{4}{20}\)
c/ \(\frac{C}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{C}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{21-20}{20.21}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{C}{2}=\frac{1}{2}-\frac{1}{21}=\frac{19}{42}\Rightarrow C=\frac{19}{21}\)
DP
14 tháng 8 2017
Đặt \(B=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{1}{210}\)
\(\frac{1}{2}B=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{420}\)
\(\frac{1}{2}B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{20.21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}B=\frac{1}{2}-\frac{1}{21}\)
\(\Rightarrow B=\frac{\frac{1}{2}-\frac{1}{21}}{\frac{1}{2}}=\frac{19}{21}\)
DP
14 tháng 8 2017
\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+50}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{\frac{\left(1+50\right).50}{2}}\)
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{1}{1275}\)
\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{2550}\)
\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{50.51}\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{50}-\frac{1}{51}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{51}\right)=2\cdot\frac{49}{102}=\frac{49}{51}\)
NP
6 tháng 5 2015
a)\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{195}\)
Đặt \(C=\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{66}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...+\frac{1}{132}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{11.12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+\left(-\frac{1}{5}+\frac{1}{5}\right)+\left(-\frac{1}{6}+\frac{1}{6}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)\(\Rightarrow\frac{1}{2}C=\frac{1}{4}+0+0+...+0-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{4}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{12}-\frac{1}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{2}{12}\)
\(\Rightarrow\frac{1}{2}C=\frac{1}{6}\)
\(\Rightarrow C=\frac{1}{6}:\frac{1}{2}\)
\(\Rightarrow C=\frac{1}{6}\cdot2\)
\(\Rightarrow C=\frac{2}{6}=\frac{1}{3}\)
A = \(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{210}\)
\(\frac{1}{2}A=\frac{1}{2}\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{210}\right)\)
\(\frac{1}{2}A=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{420}\)
\(\frac{1}{2}A=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{20.21}\)
\(\frac{1}{2}A=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{21}\)
\(\frac{1}{2}A=\frac{1}{3}-\frac{1}{21}\)
\(\frac{1}{2}A=\frac{2}{7}\)
A = \(\frac{2}{7}:\frac{1}{2}\)
A = \(\frac{4}{7}\)