\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)
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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
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2001.2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}.+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}\left(1+\frac{1}{2003}\right)\)
\(=-1+\frac{1}{2002}.\frac{2004}{2003}\)
\(=-1+\frac{2}{2003}\)
\(=\frac{-2003+2}{2003}\)
\(=\frac{-2001}{2003}\)
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 ....
MH
1
19 tháng 6 2016
\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{1.2}+\frac{1}{3.2}+...+\frac{1}{2001.2002}\right)\)
\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}\right)\)
\(=\frac{1}{2003.2002}-\left(1-\frac{1}{2002}\right)\)
\(=\frac{1}{2003.2002}-\frac{2001}{2002}\)
NT
0
HL
14 tháng 7 2017
\(\dfrac{1}{2003.2002}-\dfrac{1}{2002.2001}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
= \(\dfrac{1}{2003.2002}-\left(\dfrac{1}{2002.2001}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)
= \(\dfrac{1}{2003.2002}-\left(\dfrac{1}{2002}-\dfrac{1}{2001}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}-1\right)\)
= \(\dfrac{1}{2003.2002}-\dfrac{1}{2002}+1\)
= \(\dfrac{1-2003+2003.2002}{2003.2002}\)
= \(1-\dfrac{2002}{2003.2002}=1-\dfrac{1}{2003}\) = \(\dfrac{2002}{2003}\)
NH
1
13 tháng 9 2020
\(-\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-\frac{1}{2001\cdot2000}-...-\frac{1}{2\cdot1}\)
\(=-1\left(\frac{1}{1\cdot2}+...+\frac{1}{2000\cdot2001}+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)
\(=-1\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{2000}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(=-1\left(1-\frac{1}{2003}\right)\)
\(=-1\left(\frac{2003}{2003}-\frac{1}{2003}\right)\)
\(=-1\cdot\frac{2002}{2003}\)
\(=-\frac{2002}{2003}\)
14 tháng 3 2020
\(B=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(\Rightarrow5B=5+1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(\Rightarrow5B-B=5-\frac{1}{5^{100}}\)
\(\Rightarrow B=\frac{5-\frac{1}{5^{100}}}{4}\)
14 tháng 3 2020
\(B=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...+\frac{1}{5^{100}}\)
\(5B=1+5+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{99}}\)
\(5B-B=\left(1+5+\frac{1}{5}+...+\frac{1}{5^{99}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{100}}\right)\)
\(4B=5-\frac{1}{5^{100}}\)
\(B=\frac{5-\frac{1}{5^{100}}}{4}\)
hok tốt!!
\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{2.1}\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002.2001}+\frac{1}{2001.2000}+...+\frac{1}{2.1}\right)\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-\frac{1}{2001}+\frac{1}{2001}-\frac{1}{2000}+...+\frac{1}{2}-1\right)\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-1\right)\)
\(=\frac{1}{2003.2002}-\left(\frac{1}{2002}-\frac{2002}{2002}\right)\)
\(=\frac{1}{2003.2002}-\frac{-2001}{2002}\)
\(=\frac{1}{2003}-\frac{1}{2002}+\frac{2001}{2002}\)
\(=\frac{1}{2003}+\frac{2000}{2002}\)
\(=\frac{1}{2003}+\frac{1000}{1001}\)
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