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Khách
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MN
0
ML
9 tháng 5 2016
A=\(\frac{1}{30}\)+\(\frac{1}{42}\)+\(\frac{1}{56}\)+\(\frac{1}{72}\)+\(\frac{1}{90}\)+\(\frac{1}{110}\)+\(\frac{1}{132}\)
A=\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+\(\frac{1}{7.8}\)+\(\frac{1}{8.9}\)+\(\frac{1}{9.10}\)+\(\frac{1}{10.11}\)+\(\frac{1}{11.12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{8}\)+\(\frac{1}{8}\)-\(\frac{1}{9}\)+\(\frac{1}{9}\)-\(\frac{1}{10}\)+\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)
A= \(\frac{1}{5}\)-\(\frac{1}{12}\)=\(\frac{7}{60}\)
NT
1
MH
1
14 tháng 7 2018
Đặt \(N=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
ta có: \(M.N=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}...\frac{99}{100}.\frac{100}{101}=\frac{1}{101}\)
ta có: \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{99}{100}< \frac{100}{101}\)
\(\Rightarrow M=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}< N=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}\)
\(\Rightarrow M.M< M.N\)
\(\Rightarrow M^2< \frac{1}{101}< \frac{1}{100}=\left(\frac{1}{10}\right)^2\)
\(\Leftrightarrow M^2< \left(\frac{1}{10}\right)^2\)
\(\Rightarrow M< \frac{1}{10}\left(đpcm\right)\)
HB
0
16 tháng 6 2017
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=\left(1+\frac{1}{3}+...+\frac{1}{9}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}+\frac{1}{10}\right)-\left(1+\frac{1}{2}+...+\frac{1}{5}\right)\)
\(A=\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+...+\frac{1}{10}\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
\(B=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{5}\right)\)
Vậy A = B và A = 1/6 + 1/7 + 1/8 + 1/9 + 1/10
TH
16 tháng 6 2017
1/ A= \(\left(\frac{1}{1.2}\right)+\left(\frac{1}{3.4}\right)+...+\left(\frac{1}{9.10}\right)\)
B=(1/1+1/2+1/3+...+1/10)- (1/1+1/2+...+1/5)
<=> B=1/6+1/7+1/8+1/9+1/10.
PM
0
\(10M=1+\frac{1}{10}+\frac{1}{10^2}+...+\frac{1}{10^{2008}}\)
\(9M=10M-M=1-\frac{1}{10^{2009}}\Rightarrow M=\frac{1}{9}-\frac{1}{9.10^{2009}}< \frac{1}{9}\)