Kết quả phép tính: 1/1.3+1/3.5+1/5.7+...+1/2007.2009 help mik vs
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= 1/2 . (1/1 - 1/3 + 1/3 - 1/5 +... + 1/2009 - 1/2011)
= 1/2 . (1/1 - 1/2011)
= 1/2 . 2010 / 2011
= 1005/2011
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+............+\frac{1}{2009}-\frac{1}{2011}=\frac{1}{1}-\frac{1}{2011}=\frac{2010}{2011}\)
sai rồi top scorer ạ tử trừ mẫu là 2 mà tử là 1 phải nhân 2 lên tử
Ta có:
\(2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\right)\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{2007.2009}+\frac{2}{2009.2011}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}\)
\(\Rightarrow\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2009.2011}=\frac{2010}{2011}\div2=\frac{1005}{2011}\)
Vậy giá trị của biểu thức là \(\frac{1005}{2011}\)
a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{6}{7}\)
\(=\frac{3}{7}\)
b)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(=\frac{1}{2}.\frac{2010}{2011}\)
\(=\frac{1005}{2011}\)
2A = 2/1.3 +2/3.5 + 2/5.7 + ... + 2/2007.2009 + 2/2009. 2011
2A = 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/ 2007 - 1/2009 + 1/2009 - 1/2011
Gian uoc het ta co: 2A = 1/1 - 1/2011
2A = 2010/2011
A = 2010/2011 X 1/2
A = 1005/2011
**** mình nha
1/1*3 + 1/3*5 + 1/5*7 + ... + 1/2007*2009
= 1/2(2/1*3 + 2/3*5 + 2/5*7 + ... + 2/2007*2009)
= 1/2(1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/2007 - 1/2009)
= 1/2( 1- 1/2009)
= 1/2 * 2008/2009
= 1009/2009
2S=\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2007.2009}\)
=\(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-...+\dfrac{1}{2007}-\dfrac{1}{2009}\)
= 1- \(\dfrac{1}{2009}\)
= \(\dfrac{2008}{2009}\)
=> S=\(\dfrac{1004}{2009}\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2007}-\dfrac{1}{2009}\right)=\dfrac{1}{2}\cdot\dfrac{2008}{2009}=\dfrac{1004}{2009}\)