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2/1*2*3+2/3*4*5+...+2/2009*2010*2011
A=2/2*(1/1-1/2-1/3+1/2-1/3-1/4+1/4-1/5-1/6+...+1/2009-1/2010-1/2011
A=1*(1-1/2011)
A=1*2010/2011=2010/2011
suy ra: 2010/2011<1
suy ra 1/2 của 1 lớn hơn 2010/2011
VẬY A NHỎ HƠN 1/2
VẬY
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) \(\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\)
\(\Rightarrow100\cdot\left(\frac{1}{1\cdot101}+\frac{1}{2\cdot102}+...+\frac{1}{10\cdot110}\right)x=100\cdot\left(\frac{1}{1\cdot11}+\frac{1}{2\cdot12}+...+\frac{1}{100\cdot110}\right)\)
\(\Rightarrow\left(\frac{100}{1\cdot101}+\frac{100}{2\cdot102}+...+\frac{100}{10\cdot110}\right)x=10\cdot\left(\frac{10}{1\cdot11}+\frac{10}{2\cdot12}+...+\frac{10}{100\cdot110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{10}+\frac{1}{2}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{110}\right)\)
\(\Rightarrow\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)x=10\cdot\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+...+\frac{1}{10}-\frac{1}{110}\right)\)
\(\Rightarrow x=10\cdot\)
`A=1/(1.3)+1/(3.5)+....+1/(2009.2011)`
`=> 2A=2/(1.3)+1/(3.5)+...+2/(2009.2011)`
`=1-1/3+1/3+1/5+.....+1/2009-1/2011`
`=1-1/2011`
`=2010/2011`
`=> A=1005/2011`
Dựa trên công thức: \(\dfrac{a}{n.\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\), ta có:
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
=\(\dfrac{1}{2}.\left(1-\dfrac{1}{2011}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{2010}{2011}\right)\)
= \(\dfrac{2010}{4022}\)
= tự rút gọn nhé
Hok tốt!
Cho tổng trên là A
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
\(A=2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2003}-\frac{1}{2005}\right)\)
\(A=2\left(1-\frac{1}{2005}\right)\)
\(A=2.\frac{2004}{2005}\)
\(A=\frac{4008}{2005}\)
Đặt A = 1/1.3 + 1/3.5 + 1/5.7 +........+ 1/(2n - 1)(2n + 1)
2.A = 2/1.3 + 2/3.5 + 2/5.7 +........+ 2/(2n - 1)(2n + 1)
2.A = 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/(2n - 1) - 1/(2n + 1)
2.A = 1 - 1/(2n + 1) = 2n/(2n + 1)
Vậy A = n/(2n + 1)
hình như sai!!
\(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+\frac{4}{7\cdot9\cdot11}+\frac{4}{9\cdot11\cdot13}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)
\(=\frac{1}{1.3}-\frac{1}{11.13}\)
\(=\frac{1}{3}-\frac{1}{143}\)
\(=\frac{140}{429}\)
\(A=\frac{1}{2.3}+\frac{1}{6.5}+\frac{1}{10.7}+\frac{1}{14.9}+...+\frac{1}{198.101}\)
\(=\frac{2}{2.6}+\frac{2}{6.10}+\frac{2}{10.14}+\frac{2}{14.18}+...+\frac{2}{198.202}\)
\(=\frac{1}{2}.\left(\frac{4}{2.6}+\frac{4}{6.10}+\frac{4}{10.14}+\frac{4}{14.18}+...+\frac{4}{198.202}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+...+\frac{1}{198}-\frac{1}{202}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{202}\right)\)
\(=\frac{1}{2}.\frac{50}{101}=\frac{25}{101}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}\right)\)
\(A=2.\left(1-\frac{1}{7}\right)\)
\(A=2.\left(\frac{7}{7}-\frac{1}{7}\right)\)
\(A=2.\frac{6}{7}\)
\(A=\frac{12}{7}\)
Chúc bạn học tốt !!!
\(2.A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2007.2009}+\frac{2}{2009.2011}\)
\(2.A=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{2009-2007}{2007.2009}+\frac{2011-2009}{2009.2011}\)
\(2.A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2007}-\frac{1}{2009}+\frac{1}{2009}-\frac{1}{2011}=1-\frac{1}{2011}=\frac{2010}{2011}\)
=>A = \(\frac{2010}{2011}:2=\frac{1005}{2011}\)