Tính:
\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-\frac{1}{5.3}-\frac{1}{3.1}\)
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TZ
1
IZ
3
DH
29 tháng 1 2020
Đặt: \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{97.95}+\frac{1}{95.93}+...+\frac{1}{3.1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(1-\frac{1}{3}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{2}\left(\frac{1}{97}-\frac{1}{99}\right)-\frac{1}{2}\left(1-\frac{1}{97}\right)\)
\(=\frac{1}{2}.\frac{1}{97}-\frac{1}{2}.\frac{1}{99}-\frac{1}{2}+\frac{1}{2}.\frac{1}{97}\)
\(=-\frac{4751}{9603}\)
Vậy ....
VM
29 tháng 1 2020
\(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{97.95}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\left(1\right).\)
Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\)
\(\Rightarrow A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\)
\(\Rightarrow A=\frac{1}{2}.\left(1-\frac{1}{97}\right)\)
\(\Rightarrow A=\frac{1}{2}.\frac{96}{97}\)
\(\Rightarrow A=\frac{48}{97}.\)
+ Thay A vào \(\left(1\right)\) ta được:
\(\frac{1}{99.97}-\frac{48}{97}\)
\(=\frac{1}{99.97}-\frac{48.99}{99.97}\)
\(=\frac{1-48.99}{99.97}\)
\(=-\frac{4751}{9603}.\)
Vậy \(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}=-\frac{4751}{9603}.\)
Chúc bạn học tốt!
VK
5
GF
2
LK
28 tháng 3 2018
\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}.\)
\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{93.95}+\frac{1}{95.97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{93.95}+\frac{2}{95.97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)=\frac{1}{99.97}-\frac{1}{2}.\frac{96}{97}\)
\(=\frac{1}{99.97}-\frac{48}{97}\)
chúc bạn học tốt
KT
28 tháng 3 2018
\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-...-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{97.95}+\frac{1}{95.93}+...+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{93}-\frac{1}{95}+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{48}{97}\)
\(=-\frac{4751}{9603}\)
CT
2
15 tháng 1 2017
hình như làm nhầm r xin lỗi nha! làm lại
1/2(1/(99*97))-1/2(-1/97+1/95-1/95+1/93...+1)=1/2(1/(99*97))-1/2(-1/97+1)=-9503/19206
lần này hi vọng ko nhầm
VD
26 tháng 6 2016
đặt tổng là A
=>A=\(\frac{-1}{3}\left(\frac{1}{99}-\frac{1}{97}+\frac{1}{97}-\frac{1}{96}+....+\frac{1}{5}-\frac{1}{3}\right)\)
=>A=\(\frac{-1}{3}\left(\frac{1}{99}-\frac{1}{3}\right)=\frac{32}{297}\)
CT
A=\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-....-\frac{1}{5.3}-\frac{1}{3.1}\) giải chi tiết
0
TP
2
NN
8 tháng 1 2016
A=\(\frac{1}{99.97}\)-\(\frac{1}{97.95}\)-\(\frac{1}{95.93}\)-...-\(\frac{1}{5.3}\)-\(\frac{1}{3.1}\)
A=\(\frac{1}{2}\).(\(\frac{1}{99}\)+\(\frac{1}{97}\)-\(\frac{1}{97}\)+\(\frac{1}{95}\)-\(\frac{1}{95}\)+\(\frac{1}{93}\)-...-\(\frac{1}{5}\)+\(\frac{1}{3}\)-\(\frac{1}{3}\)+1)
A=\(\frac{1}{2}\).(\(\frac{1}{99}\)+1)
A=\(\frac{1}{2}\).\(\frac{100}{99}\)=\(\frac{50}{99}\)
8 tháng 1 2016
tớ nghĩ là bằng \(\frac{-4751}{9603}\) chứ bạn Kayoko Suzue
\(\frac{1}{99.97}-\frac{1}{97.95}-........-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.........+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.......+\frac{1}{5.3}+\frac{1}{3.1}\right).\frac{2}{2}\)
\(=-\left(-\frac{2}{99.97}+\frac{2}{97.95}+......+\frac{2}{5.3}+\frac{2}{3.1}\right).\frac{1}{2}\)
\(=-\left(-\frac{1}{99}-\frac{1}{97}+\frac{1}{97}-\frac{1}{95}+.....+\frac{1}{5}-\frac{1}{3}+\frac{1}{3}-1\right).\frac{1}{2}\)
\(=\left(\frac{1}{99}-1\right).\frac{1}{2}\)
\(=-\frac{98}{99}.\frac{1}{2}\)
\(=-\frac{49}{99}\)