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15 tháng 5 2016

sai đề

15 tháng 5 2016

\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{2009.2011}\)

\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{2009}-\frac{1}{2011}\)

\(=1-\frac{1}{2011}=\frac{2010}{2011}\)

15 tháng 11 2017

\(A=\dfrac{1}{2}+\dfrac{3-2}{3.2}+\dfrac{4-3}{3.4}+...+\dfrac{100-99}{100.99}\)

\(A=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=1-\dfrac{1}{100}\)

\(A=\dfrac{99}{100}\)

15 tháng 11 2017

\(2B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+....+\dfrac{2}{2007.2009}+\dfrac{2}{2009..2011}\)

\(2B=\dfrac{3-1}{1.3}+\dfrac{5-3}{3,5}+...+\dfrac{2009-2007}{2009.2007}+\dfrac{2011-2009}{2011.2009}\)

\(2B=\dfrac{3}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2011}\)

\(2B=1-\dfrac{1}{2011}\)

\(2B=\dfrac{2010}{2011}\)

\(B=\dfrac{2010}{4022}\)

25 tháng 1 2016

Ta có:

\(S=\frac{4}{5.7}+\frac{4}{7.9}+\frac{4}{9.11}+...+\frac{4}{59.61}\)

\(=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(=2.\left(\frac{61}{305}-\frac{5}{305}\right)\)

\(=2.\frac{56}{305}\)

\(=\frac{112}{305}\)

Vậy \(S=\frac{112}{305}\)

22 tháng 9 2016

\(\Rightarrow A=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{59.61}\right)\)

\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{59}-\frac{1}{61}\right)\)

\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(\Rightarrow A=\frac{3}{2}.\frac{56}{305}\)

\(\Rightarrow A=\frac{84}{305}\)

13 tháng 7 2017

\(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{19.21}\)

\(=2\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{19.21}\right)\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(=2\left(1-\dfrac{1}{21}\right)=2.\dfrac{20}{21}=\dfrac{40}{21}\)