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Đặt \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(\Rightarrow A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2009}-\frac{1}{2010}\)
\(\Rightarrow A=1-\frac{1}{2010}=\frac{2010}{2010}-\frac{1}{2010}=\frac{2009}{2010}\)
Vậy \(A=\frac{2009}{2010}\)
1/1*2+1/2*3+........+1/2009*2010
=1-1/2+1/2-1/3+..........+1/2009-1/2010
=1-1/2010
=2009/2010
*2010/1+2009/2+...+1/2010
=(2009/2+1)+(2008/3+1)+...+(1/2010+1)+1
=2011/2+2011/3+..+2011/2010+2011/2011
=2011(1/2+1/3+1/4+...+1/2011)
=> C=2011/1=2011
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2010}\right)\)
\(=\left(\dfrac{2-1}{2}\right)\left(\dfrac{3-1}{3}\right)\left(\dfrac{4-1}{4}\right)...\left(\dfrac{2010-1}{2010}\right)\)
\(=\dfrac{1.2.3...2009}{2.3.4...2010}\)
\(=\dfrac{1}{2010}\)
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\dots\left(1-\dfrac{1}{2009}\right)\cdot\left(1-\dfrac{1}{2010}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\cdot\cdot\dfrac{2008}{2009}\cdot\dfrac{2009}{2010}\)
\(=\dfrac{1\cdot2\cdot3\dots2008\cdot2009}{2\cdot3\cdot4\dots2009\cdot2010}\)
\(=\dfrac{1}{2010}\)
Chào mai xinh đẹp
1<=>( x-4)/2009 -1 +( x-3)/2010-1 -(x-2)/2011-1-(x-1)/2012-1=0
<=> (x-2013)/2009+ (x-2013)/2010-(x-2013)/2011-(x-2013)/2012=0
<=> (x-2013)( 1/2009+1/2010-1/2011-1/2012)=0
=> x-2013=0=> x=2013
pp mai
Lời giải:
\(F=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2010}{2009}.\frac{2011}{2010}\\ =\frac{3.4.5...2010.2011}{2.3.4...2009.2010}=\frac{2011}{2}\)