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Khách
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17 tháng 7 2019
ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
tách
\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\frac{1}{2}-\frac{1}{1024}\)
thay vào B ta có
\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
17 tháng 7 2019
\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)
\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)
\(\Rightarrow A=\frac{513}{1024}\)
HN
16 tháng 9 2017
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.............-\frac{1}{1024}\)
=> 2S = \(2x\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..........-\frac{1}{1024}\right)\)
2S = \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..........-\frac{1}{512}\)
2S - S = \(\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-........-\frac{1}{512}\right)\)- \(\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-........-\frac{1}{1024}\right)\)
=> S = \(1+\frac{1}{1024}=\frac{1024}{1024}+\frac{1}{1024}=\frac{1025}{1024}\)
Chắc chắn 100%
TN
0
FY
1
20 tháng 10 2019
https://olm.vn/hoi-dap/tim-kiem?q=T%C3%ACm+x,+bi%E1%BA%BFt:+3x2.5++3x5.8++3x8.11++3x11.14+=121+&id=81551
Cậu vào link này nhé(đây là đáp án câu này)
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−\(\frac{1}{97}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).\(\frac{96}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48}{97}\)
=\(\frac{1}{99.97}\)−\(\frac{48.99}{99.97}\)
=\(\frac{-4751}{9603}\)
UN
2
14 tháng 3 2017
\(\dfrac{1}{99.97}-\dfrac{1}{97.95}-\dfrac{1}{95.93}-...-\dfrac{1}{5.3}-\dfrac{1}{3.1}\)
=\(\dfrac{1}{99.97}-\)(\(\dfrac{1}{97.95}+\dfrac{1}{95.93}+...+\dfrac{1}{5.3}+\dfrac{1}{3.1}\))
=\(\dfrac{1}{99.97}-\)\(\dfrac{1}{2}\left(\dfrac{1}{95}-\dfrac{1}{97}+\dfrac{1}{93}-\dfrac{1}{95}+\dfrac{1}{3}-\dfrac{1}{5}+1-\dfrac{1}{3}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}\left(1-\dfrac{1}{97}\right)\)
=\(\dfrac{1}{99.97}-\dfrac{1}{2}.\dfrac{96}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48}{97}\)
=\(\dfrac{1}{99.97}-\dfrac{48.99}{99.97}\)
=\(\dfrac{-4751}{9603}\)
Ta có : \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-.....-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\)
=> \(2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{512}\)
=> \(2A-A=\frac{1}{2}-\frac{1}{1024}\)
Thay A vào ta có : \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
Jenny123 tham khảo nhé
Đặt tổng trên là A, ta có:
\(A.2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(A.2-A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{512}-"\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)
\(\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}"\)
\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}-\frac{1}{128}-\frac{1}{256}-\frac{1}{512}-\frac{1}{1024}\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
P/s: Bn xem lại đề nha