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1/1x2 + 1/2x3+ 1/3x4+....+1/2009x2010
= 1/1-1/2 + 1/2-1/3+ 1/3-1/4+...+1/2009-1/2010
= 1/1-1/2010
= 2009/2010
\(\frac{1}{1\cdot2}+\cdot\cdot\cdot+\frac{1}{2009\cdot2010}\)
\(=1-\frac{1}{2}+\cdot\cdot\cdot+\frac{1}{2009}-\frac{1}{2010}\)
\(=1-\frac{1}{2010}\)
\(=\frac{2009}{2010}\)
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Hình như đề bài phải là : Tính tổng : \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}+\frac{1}{2010.2011}\)
Nếu thế giải như sau : \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}+\frac{1}{2010}-\frac{1}{2011}\)
\(=1-\frac{1}{2011}=\frac{2010}{2011}.\)Vậy tổng đó là 2010/2011.
Ta có :\(\frac{1}{1}:2+\frac{1}{2}:3+...+\frac{1}{2010}:2011\)
= \(\frac{1}{1}\times\frac{1}{2}+\frac{1}{2}\times\frac{1}{3}+...+\frac{1}{2010}\times\frac{1}{2011}\)
= \(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{2010\times2011}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2010}-\frac{1}{2011}\)
= \(1-\frac{1}{2011}\)
= \(\frac{2010}{2011}\)
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đầu bài là ê các bạn mình thấy lạ thật đấy chưa từng thấy đầu bài nào kì cục như vậy
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=(1-2-3+4)+(5-6-7+8)+...+(2005-2006-2007+2008)+2009
=2009
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\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{x.\left(x+2\right)}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{20}{41}:\frac{1}{2}\)
\(\Rightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Rightarrow\frac{1}{x+2}=1-\frac{40}{41}=\frac{1}{41}\)
=> x + 2 = 41
=> x = 39
`#P.Vy`
`C=(3/3+2/3)(5/5+2/5)(7/7+2/7)...(2009/2009+2/2009)(2011/2011+2/2011)`
`C=5/3xx7/5xx9/7xx...xx2011/2009xx2013/2011`
\(C=\dfrac{5.7.9\times...\times2011.2013}{3.5.7\times...\times2009.2011}\)
`C=671`
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