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![](https://rs.olm.vn/images/avt/0.png?1311)
*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
![](https://rs.olm.vn/images/avt/0.png?1311)
a,S1=1+(-2)+3+(-4)+..........+2009+(-2010)
S1=-1.(2010:2)
S1=-1005
b,S2=1+(-2)+(-3)+4+5+(-6)+(-7)+............+2008+2009+(-2010)
S2=-1.(2010:2)
S2=-1.1005
S2=-1005
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2010}\right)\)
\(=\left(\dfrac{2-1}{2}\right)\left(\dfrac{3-1}{3}\right)\left(\dfrac{4-1}{4}\right)...\left(\dfrac{2010-1}{2010}\right)\)
\(=\dfrac{1.2.3...2009}{2.3.4...2010}\)
\(=\dfrac{1}{2010}\)
\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\dots\left(1-\dfrac{1}{2009}\right)\cdot\left(1-\dfrac{1}{2010}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\cdot\cdot\dfrac{2008}{2009}\cdot\dfrac{2009}{2010}\)
\(=\dfrac{1\cdot2\cdot3\dots2008\cdot2009}{2\cdot3\cdot4\dots2009\cdot2010}\)
\(=\dfrac{1}{2010}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Kết quả : 0
Giải:
(-2012+2012)+(-2011+2011)+(-2010+2010)+(-2009+2009)+................+(-3+3)+(-2+2)+(-1+1)+0=0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
cho hỏi chỗ 1-1/2009 và 1-1/2010 là dấu cộng hay trừ để mik bt lối ạ