Tính:
1 / 2003.2004 - 1 / 2002.2003 - . . . - 1 / 3.2 - 1 / 2.1
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12 tháng 4 2019
\(=\frac{1}{1.2}-\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}\right)\)
\(=\frac{1}{1.2}-\left(\frac{1}{2}-\frac{1}{99}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{99}\)
\(=\frac{1}{99}\)
PT
0
NT
5
HT
17 tháng 7 2015
\(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-....-\frac{1}{2.1}\)
=\(-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\right)\)
=\(-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\right)\)
=\(-\left(1-\frac{1}{100}\right)\)
=\(\frac{-99}{100}\)
17 tháng 7 2015
A=1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
A= - (1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
A= - (1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
A= - (1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
A= - (1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
A= - (1/1-1/100)
A= - 99/100
26 tháng 6 2017
a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(A=1-\frac{1}{99}\)
\(A=\frac{98}{99}\)
thay A vào, ta được :
\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)
\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)
đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(A=2.\left(1-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Thay A vào, ta được :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
27 tháng 8 2015
à mình nhầm có phải thế này không
1/100.99 - 1/99.98 - 1/98.97 -...- 1/3.2 - 1/2.1
=-(1/100.99 + 1/99.98 + 1/98.97 +...+ 1/3.2 + 1/2.1)
=-(1/2.1+1/3.2 +...+1/98.97+ 1/99.98 +1/100.99 )
=-(1/1.2+1/2.3+1/3.4+...+1/97.98+ 1/98.99 +1/99.100)
=-(1/1-1/2+1/2-1/3+1/3......-1/98+1/98-1/99+1/99-1/100)
=-(1/1-1/100)=-99/100
LQ
2
NV
14 tháng 7 2018
\(P=\)\(-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2018}+\frac{1}{2018}\)
\(P=0\)
14 tháng 7 2018
\(P=-1+\frac{1}{2.1}+\frac{1}{3.2}+\frac{1}{4.3}+...+\frac{1}{2018.2017}+\frac{1}{2018}\)
\(P=-1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}+\frac{1}{2018}\)
\(P=-1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}+\frac{1}{2018}\)
P = 0
TT
1
NN
1
TL
0
Đặt \(A=\frac{1}{2003.2004}-\frac{1}{2002.2003}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow-A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2002.2003}+\frac{1}{2003.2004}\)
\(\Rightarrow-A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2003}-\frac{1}{2004}\)
\(\Rightarrow-A=1-\frac{1}{2004}\)
\(\Rightarrow-A=\frac{2003}{2004}\)
\(\Rightarrow A=\frac{-2003}{2004}\)