Tính hợp lí:
A=\(\frac{1}{299.297}\)- \(\frac{1}{297.295}\) - \(\frac{1}{295.293}\)-...-\(\frac{1}{3.1}\)
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DD
1
4 tháng 3 2023
`A=1/(299.297)-1/(297.295)-1/(295.293)-...-1/(3.1)`
`2A=2/(299.297)-(2/(297.295)+2/(295.293)+....+2/(3.1))`
`2A=2/(299.297)-(1/295-1/297+1/293-1/295+....+1-1/3)`
`2A=2/(299.297)-(1-1/297)`
`2A=2/(299.297)-296/297`
`A=1/(299.297)-148/297=(1-148.299)/(299.297)`
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\left( { - \frac{5}{6}} \right) - \left( { - 1,8} \right) + \left( { - \frac{1}{6}} \right) - 0,8\\ = \left( { - \frac{5}{6}} \right) + 1,8 + \left( { - \frac{1}{6}} \right) - 0,8\\ = \left[ {\left( { - \frac{5}{6}} \right) + \left( { - \frac{1}{6}} \right)} \right] + \left[ {1,8 - 0,8} \right]\\ =\frac{-6}{6}+1= - 1 + 1 = 0\end{array}\)
b)
\(\begin{array}{l}\left( { - \frac{9}{7}} \right) + \left( { - 1,23} \right) - \left( { - \frac{2}{7}} \right) - 0,77\\ = \left[ {\left( { - \frac{9}{7}} \right) - \left( { - \frac{2}{7}} \right)} \right] + \left[ {\left( { - 1,23} \right) - 0,77} \right]\\ =\frac{-7}{7}+(-2)= - 1 + \left( { - 2} \right) = - 3\end{array}\)
TZ
1
DV
1
N
22 tháng 7 2017
\(\frac{1}{99.97}-\frac{1}{97.95}-........-\frac{1}{5.3}-\frac{1}{3.1}\)
\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.........+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
\(=-\left(-\frac{1}{99.97}+\frac{1}{97.95}+.......+\frac{1}{5.3}+\frac{1}{3.1}\right).\frac{2}{2}\)
\(=-\left(-\frac{2}{99.97}+\frac{2}{97.95}+......+\frac{2}{5.3}+\frac{2}{3.1}\right).\frac{1}{2}\)
\(=-\left(-\frac{1}{99}-\frac{1}{97}+\frac{1}{97}-\frac{1}{95}+.....+\frac{1}{5}-\frac{1}{3}+\frac{1}{3}-1\right).\frac{1}{2}\)
\(=\left(\frac{1}{99}-1\right).\frac{1}{2}\)
\(=-\frac{98}{99}.\frac{1}{2}\)
\(=-\frac{49}{99}\)
LT
1
XO
23 tháng 8 2019
\(\frac{1}{99.97}-\frac{1}{97.95}-...-\frac{1}{3.1}\)
\(=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{95.97}\right)\)
\(=\frac{1}{2}.\frac{2}{97.99}-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\)
\(=\frac{1}{2}.\left[\frac{2}{97.99}-\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{95.97}\right)\right]\)
\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{95}-\frac{1}{97}\right)\right]\)
\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\left(1-\frac{1}{97}\right)\right]\)
\(=\frac{1}{2}.\left(\frac{1}{97}-\frac{1}{99}-\frac{98}{97}\right)\)
\(=\frac{1}{2}.\left(-1-\frac{1}{99}\right)\)
\(=\frac{1}{2}.\frac{-100}{99}\)
\(=-\frac{50}{99}\)
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=\(\frac{-1}{3}\left(\frac{1}{99}-\frac{1}{3}\right)=\frac{32}{297}\)
13 tháng 8 2017
Tôi thấy bài này nó cứ sai sai
Ở chỗ \(\frac{1}{99.97}-\frac{1}{97.95}\)í
\(\frac{1}{97.95}>\frac{1}{99.97}\)mà ông Thám Tử THCS Nguyễn Hiếu CTV
13 tháng 8 2017
violympic cho sai đề :
Đề đúng là tính : \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.53}-....-\frac{1}{5.3}-\frac{1}{3.1}\)
Làm theo đề đúng !! ok
Ta có : \(A=\frac{1}{99.97}-\left(\frac{1}{97.95}+\frac{1}{95.53}+....+\frac{1}{5.3}+\frac{1}{3.1}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{95}-\frac{1}{97}\right)\)
\(=\frac{1}{99.97}-\frac{1}{2}\left(1-\frac{1}{97}\right)=\frac{1}{99.97}-\frac{48}{97}=-\frac{4751}{9603}\)
LA
15 tháng 3 2017
a)\(\frac{1}{99.97}\)−\(\frac{1}{97.95}\)−\(\frac{1}{95.93}\)−…−\(\frac{1}{5.3}\)−\(\frac{1}{3.1}\)
=\(\frac{1}{99.97}\)−(\(\frac{1}{97.95}\)+\(\frac{1}{95.93}\)+…+\(\frac{1}{5.3}\)+\(\frac{1}{3.1}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(\(\frac{1}{95}\)−\(\frac{1}{97}\)+\(\frac{1}{93}\)−\(\frac{1}{95}\)+…+\(\frac{1}{3}\)−\(\frac{1}{5}\)+1−\(\frac{1}{3}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(1−\(\frac{1}{97}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).\(\frac{96}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48.99}{99.97}\)
=\(\frac{-4751}{9603}\)
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a)
\(\begin{array}{l}\frac{{ - 3}}{{10}} - 0,125 + \frac{{ - 7}}{{10}} + 1,125 \\= \left( {\frac{{ - 3}}{{10}} + \frac{{ - 7}}{{10}}} \right) + \left( {1,125 - 0,125} \right)\\ = - 1 + 1 \\= 0\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 8}}{3}.\frac{2}{{11}} - \frac{8}{3}:\frac{{11}}{9} \\= \frac{8}{3}.\frac{{ - 2}}{{11}} - \frac{8}{3}.\frac{9}{{11}}\\ = \frac{8}{3}.\left( {\frac{{ - 2}}{{11}} - \frac{9}{{11}}} \right)\\ =\frac{{ - 8}}{3}.\frac{-11}{11}\\= \frac{8}{3}.\left( { - 1} \right) \\= \frac{{ - 8}}{3}\end{array}\)