\(\frac{1}{3}m^2+\frac{1}{3}m^2+1\frac{1}{3}m^2=........m^2\)
K
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TK
2
VN
11 tháng 12 2016
M = \(\frac{1}{3}\)+ \(\frac{1}{6}\)+ \(\frac{1}{10}\)+ \(\frac{1}{15}\)( Mẫu chung: 60 )
M = \(\frac{20}{60}\)+ \(\frac{10}{60}\)+ \(\frac{6}{60}\)+ \(\frac{4}{60}\)
M = \(\frac{40}{60}\)
M = \(\frac{2}{3}\)
N
5 tháng 7 2019
\(\frac{1}{M}=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+...+\frac{1}{\frac{59.60}{2}}\)
\(\frac{1}{M}=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{59.60}\)
\(\frac{1}{M}=2.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....+\frac{1}{59}-\frac{1}{60}\right)\)
\(\frac{1}{M}=\frac{2}{3}-\frac{2}{60}< \frac{2}{3}\)
-theo t đề là M chứ ko phải 1/M
19 tháng 4 2020
Ta có : \(\frac{a^3-1}{\left(a+1\right)^3+1}=\frac{\left(a-1\right)\left(a^2+a+1\right)}{\left(a+1+1\right)\left(\left(a+1\right)^2-\left(a+1\right)+1\right)}=\frac{a-1}{a+2}\)
\(M=\frac{100^3-1}{2^3+1}.\frac{2^3-1}{3^3+1}.\frac{3^3-1}{4^3+1}...\frac{99^3-1}{100^3+1}\)
\(M=\frac{999999}{9}.\frac{1}{4}.\frac{2}{5}.\frac{3}{6}...\frac{98}{101}=\frac{999999.1.2.3}{9.99.100.101}\)
\(M=\frac{10101.2}{3.100.101}=\frac{20202}{30300}>\frac{20200}{30300}=\frac{2}{3}\)
KB
24 tháng 3 2019
\(A=\frac{m-1}{1}+\frac{m-2}{2}+...+\frac{2}{m-2}+\frac{1}{m-1}\)
\(=\frac{m-1}{1}+\frac{m-2}{2}+...+\frac{m-\left(m-2\right)}{m-2}+\frac{m-\left(m-1\right)}{m-1}\)
\(=m+\frac{m}{2}+\frac{m}{3}+...+\frac{m}{m-1}-1-1-...-1\)
\(=m+\frac{m}{2}+\frac{m}{3}+...+\frac{m}{m-1}-\left(m-1\right)\)
\(=\frac{m}{2}+\frac{m}{3}+...+\frac{m}{m-1}+\frac{m}{m}\)
\(=m\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{m}\right)\)
\(\Rightarrow\frac{A}{B}=m\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
29 tháng 6 2020
\(M=\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{4005}\)
\(\frac{M}{2}=\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{8010}\)
\(\frac{M}{2}=\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{89x90}\)
\(\frac{M}{2}=\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}+...+\frac{90-89}{89.90}\)
\(\frac{M}{2}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{89}-\frac{1}{90}=\frac{1}{3}-\frac{1}{90}\)
\(M=\frac{2}{3}-\frac{2}{90}< \frac{2}{3}\)
\(\frac{1}{3}m^2+\frac{1}{3}m^2+1\frac{1}{3}m^2=\frac{2}{3}m^2+\frac{4}{3}m^2=\frac{6}{3}m^2=2m^2\)
1/3m2+1/3m2+4/3m2=2m2