so sánh \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{205^2}v\text{à}\frac{1}{2^2\cdot3\cdot5^2\cdot7}\)
K
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ND
1
17 tháng 3 2016
2/1.3+2/3.5+2/5.7+...+2/101.103=1/1-1/3+1/3-1/5+1/5-1/7+...+1/101-1/103
=1-1/103=102/103
k cho mình nhé chắc chắn đúng
1 tháng 8 2018
\(=\frac{219}{520}=\frac{155052}{368160}\)
\(=\frac{303}{708}=\frac{157560}{368160}\)
\(\frac{155052}{368160}< \frac{157560}{368160}\)
VẬY \(\frac{303}{708}\)LỚN HƠN
AK
7 tháng 5 2018
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(=1-\frac{1}{11}\)
\(=\frac{11}{11}-\frac{1}{11}\)
\(=\frac{10}{11}\)
Chúc bạn học tốt !!!
Q
2
28 tháng 6 2016
\(E=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\)
\(E=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(E=\frac{1}{1}-\frac{1}{99}\)
\(E=\frac{98}{99}\)
28 tháng 6 2016
E= \(\frac{2}{1.3}.\frac{2}{3.5}+...+\frac{2}{97.99}\)
E = 1 - \(\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
E = 1 - 1/99
E = 98 / 99
Chúc bạn học tốt
3 tháng 5 2018
a) \(A=\frac{1}{1\cdot3\cdot5}+\frac{1}{3\cdot5\cdot7}+...+\frac{1}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+...+\frac{4}{25\cdot27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{25\cdot27}-\frac{1}{27\cdot29}\)
\(\Rightarrow4A=\frac{1}{1\cdot3}-\frac{1}{27\cdot29}=\frac{1}{3}-\frac{1}{783}=\frac{261}{783}-\frac{1}{783}=\frac{260}{783}\)
\(\Rightarrow A=\frac{\frac{260}{783}}{4}=\frac{65}{783}\)
b) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
30 tháng 4 2018
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
AK
30 tháng 4 2018
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\left(\frac{7}{7}-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
Chúc bạn học tốt !!!
9 tháng 5 2018
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+\frac{1}{14.9}+...+\frac{1}{198.101}\)
\(=\frac{2}{2.6}+\frac{2}{6.10}+\frac{2}{10.14}+\frac{2}{14.18}+...+\frac{2}{198.202}\)
\(=\frac{1}{2}.\left(\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+\frac{4}{14.18}+...+\frac{4}{198.202}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(=\frac{1}{2}.\frac{50}{101}=\frac{25}{101}\)
Đặt A = \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{205^2}\)
=> A < \(\frac{1}{100.101}+\frac{1}{101.102}+....+\frac{1}{204.205}\)
=> A < \(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+...+\frac{1}{204}-\frac{1}{205}\)
=> A < \(\frac{1}{100}-\frac{1}{205}\)
=> A < \(\frac{1}{2100}\)
Đặt B = \(\frac{1}{2^2.3.5^2.7}=\frac{1}{2100}\)
=> A < B
=> \(\frac{1}{101^2}+\frac{1}{102^2}+...+\frac{1}{205^2}
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