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11 tháng 9 2016

\(\Rightarrow\frac{M}{512}=1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\)

\(\Rightarrow2.\left(\frac{M}{512}\right)=2-1-\frac{1}{2}-.....-\frac{1}{2^9}\)

\(\Rightarrow2.\left(\frac{M}{512}\right)-\frac{M}{512}=\left(2-1-\frac{1}{2}-.....-\frac{1}{2^9}\right)-\left(1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\right)\)

\(\Rightarrow\frac{M}{512}=-\frac{1}{2^{10}}\)

\(\Rightarrow M=-\frac{1}{2}\)

 

29 tháng 9 2015

M= 512 - \(\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)

=> 2.M = 1024  - 512 -  \(\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^9}\)

=> 2.M - M = 1024 - 512 - 512 + \(\frac{512}{2^{10}}\)

=> M = \(\frac{512}{2^{10}}=\frac{2^9}{2^{10}}=\frac{1}{2}\)

29 tháng 9 2015

M = \(512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-.....-\frac{512}{2^{10}}\)

M = \(512-512.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

Đặt A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

2A = \(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{11}}\)

A = 2A - A = \(1-\frac{1}{2^{10}}\)

=> M = \(512-512.\left(1-\frac{1}{2^{10}}\right)\)

=> M = 512.\(\left(1-1+\frac{1}{2^{10}}\right)\)

=> M = \(512.\frac{1}{2^{10}}\)

=> M = \(\frac{512}{2^{10}}\)

24 tháng 12 2021

Tham khảo:Câu hỏi của Nguyễn Thị Thanh Bình - Toán lớp 7 - Học trực tuyến OLM

Ta có:

\(P=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)

\(\Rightarrow P=512-\left(\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+...+\frac{512}{2^{10}}\right)\)

Đặt \(A=\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+...+\frac{512}{2^{10}}\)

\(\Rightarrow2A=512+\frac{512}{2}+\frac{512}{2^2}+...+\frac{512}{2^9}\)

\(\Rightarrow2A-A=512-\frac{512}{2^{10}}\)

\(\Rightarrow A=512-\frac{512}{2^{10}}\)

\(\Rightarrow P=512-A=512-\left(512-\frac{512}{2^{10}}\right)=\frac{512}{2^{10}}=\frac{1}{2}\)

12 tháng 2 2020

\(P=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)

\(\Rightarrow P=2^9-\frac{2^9}{2}-\frac{2^9}{2^2}-\frac{2^9}{2^3}-...-\frac{2^9}{2^{10}}\)

\(\Rightarrow P=2^9-2^8-2^7-2^6-...-\frac{1}{2}\)

\(\Rightarrow2P=2^{10}-2^9-2^8-2^7-...-1\)

\(\Rightarrow2P-P=2^{10}-2^9-2^8-2^7-...-1-\left(2^9-2^8-2^7-2^6-...-\frac{1}{2}\right)\)

\(\Rightarrow2P-P=2^{10}-2^9-2^8-2^7-...-1-2^9+2^8+2^7+2^6+...+\frac{1}{2}\)

\(\Rightarrow P=2^{10}-2^9-2^9+\frac{1}{2}\)

\(\Rightarrow P=2^{10}-2.2^9+\frac{1}{2}\)

\(\Rightarrow P=2^{10}-2^{10}+\frac{1}{2}\)

\(\Rightarrow P=0+\frac{1}{2}\)

\(\Rightarrow P=\frac{1}{2}.\)

Chúc bạn học tốt!

28 tháng 2 2020

\(B=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\\ =512\cdot\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{10}}\right)\\ =512\cdot\left[1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\right]\)

Đặt \(H=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\Leftrightarrow B=512\cdot\left(1-H\right)\)

\(\Leftrightarrow2H=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\\ \Leftrightarrow2H-H=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\\ \Leftrightarrow H=1-\frac{1}{2^{10}}\\ \Leftrightarrow B=512\cdot\left[1-\left(1-\frac{1}{2^{10}}\right)\right]\\ \Leftrightarrow B=512\cdot\frac{1}{2^{10}}\\ \Rightarrow B=2^9\cdot\frac{1}{2^{10}}\\ \Rightarrow B=\frac{1}{2}\)

9 tháng 7 2015

M = 512 - 512/2 - .... - 512/2^10

   = 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10

   = 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2

2M = 2^10 - 2^9 - 2^8 - .... - 1 

2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9  + 2^8 + 2^7 +... +    1 + 1/2

          M   = 2^10 - 2.2^9 + 1/2

          M  = 2^10 - 2^10 + 1/2

          M  = 1/2

19 tháng 1 2018

M = 512 - 512/2 - .... - 512/2^10
   = 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
   = 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1 
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9  + 2^8 + 2^7 +... +    1 + 1/2
          M   = 2^10 - 2.2^9 + 1/2
          M  = 2^10 - 2^10 + 1/2
          M  = 1/2