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14 tháng 7 2018
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-\frac{1}{2001.2000}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{2003.2002}+\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
\(=-\left(\frac{1}{2003}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2001}+...+\frac{1}{3}-\frac{1}{2}+\frac{1}{2}-1\right)\)
\(=-\left(\frac{1}{2003}-1\right)=-\left(-\frac{2002}{2003}\right)=\frac{2002}{2003}\)
Vậy ....
HL
14 tháng 7 2017
\(\dfrac{1}{2003.2002}-\dfrac{1}{2002.2001}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
= \(\dfrac{1}{2003.2002}-\left(\dfrac{1}{2002.2001}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
= \(\dfrac{1}{2003.2002}-\left(\dfrac{1}{2002}-\dfrac{1}{2001}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\right)\)
= \(\dfrac{1}{2003.2002}-\dfrac{1}{2002}+1\)
= \(\dfrac{1-2003+2003.2002}{2003.2002}\)
= \(1-\dfrac{2002}{2003.2002}=1-\dfrac{1}{2003}\) = \(\dfrac{2002}{2003}\)
L
30 tháng 8 2019
Ta có :
\(A=\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(A=-\left(\frac{1}{2003\cdot2002}+\frac{1}{2002\cdot2001}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(A=-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)
\(A=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(A=-\left(1-\frac{1}{2003}\right)\)
\(A=-\frac{2002}{2003}\)
30 tháng 8 2019
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2001.2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}.+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}\left(1+\frac{1}{2003}\right)\)
\(=-1+\frac{1}{2002}.\frac{2004}{2003}\)
\(=-1+\frac{2}{2003}\)
\(=\frac{-2003+2}{2003}\)
\(=\frac{-2001}{2003}\)
NT
1
TG
13 tháng 9 2020
\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{2.1}\right)\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2000}+...+\frac{1}{2}-1\right)\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-1\right)\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-\frac{2002}{2002}\right)\)
\(=\frac{1}{2003.2002}-\frac{-2001}{2002}\)
\(=\frac{1}{2003}-\frac{1}{2002}+\frac{2001}{2002}\)
\(=\frac{1}{2003}+\frac{2000}{2002}\)
\(=\frac{1}{2003}+\frac{1000}{1001}\)
Ko chắc ạ! Bạn nào rãnh thì check hoặc nhận xét hộ!
\(-\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-\frac{1}{2001\cdot2000}-...-\frac{1}{2\cdot1}\)
\(=-1\left(\frac{1}{1\cdot2}+...+\frac{1}{2000\cdot2001}+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)
\(=-1\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=-1\left(1-\frac{1}{2003}\right)\)
\(=-1\left(\frac{2003}{2003}-\frac{1}{2003}\right)\)
\(=-1\cdot\frac{2002}{2003}\)
\(=-\frac{2002}{2003}\)