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gọi biểu thức đó là A
\(A=\frac{1}{3.5}+\frac{1}{5.7}+.......+\frac{1}{2009.2011}\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.......+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{2008}{6033}\right)\)
\(A=\frac{1004}{6033}\)
mink nghĩ vậy bạn ạ
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)
\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)
\(=\frac{1}{3}-\frac{1}{47.49}\)
\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)
=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))
=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))
=>A=\(\frac{350}{101}\)
7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))
7/2 ( 1 - 1/101 )
7/2 x 100/101
=350/101
\(\frac{1}{1x3x5}+\frac{1}{5x7x9}+\frac{1}{9x11x13}+.....+\frac{1}{49x51x53}=\)
\(1-\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{49}-\frac{1}{51}-\frac{1}{53}=\)
\(1-\frac{1}{3}-\frac{1}{7}-....-\frac{1}{51}-\frac{1}{53}=\)
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10
=1/2-1/10
=5/10-1/10
=4/10=2/5
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}\)
\(\frac{2}{5}\)
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(1-\frac{1}{101}\right)\)
\(=2.\frac{100}{101}=\frac{200}{101}\)
Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+..+\frac{4}{99.101}\)
\(A=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(A=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=2.\left(1-\frac{1}{101}\right)\)
\(A=\frac{2.100}{101}=\frac{200}{101}\)
Ủng hộ mk nha !!! ^_^
\(\frac{4}{1.3}+\frac{4}{3.5}+........+\frac{4}{2011.2013}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+........+\frac{2}{2011.2013}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+........+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=2\left(1-\frac{1}{2013}\right)\)
\(=2.\frac{2012}{2013}\)
\(=\frac{4024}{2013}\)
Bài làm
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{x.\left(x+2\right)}=\frac{2015}{2016}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+2}=\frac{2015}{2016}\)
\(1-\frac{1}{x+2}=\frac{2015}{2016}\)
\(\frac{1}{x+2}=\frac{1}{2016}\)
\(\Rightarrow x+2=2016\)
\(x=2014\)
Ta có:\(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2009.2011}\)
\(2A=2.\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2009.2011}\right)=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2009.2011}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
=>\(A=\frac{2010}{2011}:2=\frac{2010}{2011}.\frac{1}{2}=\frac{1005}{2011}\)