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Tinh nhanh : hihi Help me !!!

A= \(\dfrac{1}{1.3}\) + \(\dfrac{1}{3.5}\) + \(\dfrac{1}{5.7}\) + ... + \(\dfrac{1}{2009.2011}\)

TT
6 tháng 5 2017 lúc 16:57

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)

\(A=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{2011}\right)\)

\(A=\dfrac{1}{2}.\dfrac{2010}{2011}\)

\(A=\dfrac{1005}{2011}\)

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NT
6 tháng 5 2017 lúc 16:58

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{2009.2011}\)

\(=\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2009.2011}\right)\)

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

\(=\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right)\)

\(=\dfrac{1}{2}.\dfrac{2010}{2011}\)

\(=\dfrac{1005}{2011}\)

Vậy \(A=\dfrac{1005}{2011}\)

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NP
6 tháng 5 2017 lúc 16:48

A=1.(1-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\)\(\dfrac{1}{2011}\))

A=1.(1-\(\dfrac{1}{2011}\))

A=1.(\(\dfrac{2011-1}{2011}\))

A=1.\(\dfrac{2010}{2011}\)

A=\(\dfrac{2010}{2011}\)

Chúc bạn học tốt nha!

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MN
6 tháng 5 2017 lúc 18:32

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2009.2011}\)

=\(\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2009.2011}\right)\)

=\(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)

= \(\dfrac{1}{2}.\left(\dfrac{1}{1}-\dfrac{1}{2011}\right)\)

=\(\dfrac{1}{2}.\dfrac{2010}{2011}\)

=\(\dfrac{1005}{2011}\)

Vậy A=\(\dfrac{1005}{2011}\)

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