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HT
1
IM
11 tháng 9 2016
\(\Rightarrow\frac{M}{512}=1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\)
\(\Rightarrow2.\left(\frac{M}{512}\right)=2-1-\frac{1}{2}-.....-\frac{1}{2^9}\)
\(\Rightarrow2.\left(\frac{M}{512}\right)-\frac{M}{512}=\left(2-1-\frac{1}{2}-.....-\frac{1}{2^9}\right)-\left(1-\frac{1}{2}-\frac{1}{2^2}-.....-\frac{1}{2^{10}}\right)\)
\(\Rightarrow\frac{M}{512}=-\frac{1}{2^{10}}\)
\(\Rightarrow M=-\frac{1}{2}\)
L
7
ND
3
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 = \(\frac{512}{2^{10}}=\frac{2^9}{2^{10}}=\frac{1}{2}\)
HT
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 = \(\frac{512}{2^{10}}\)
NT
0
HP
3 tháng 7 2016
\(512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-......-\frac{512}{2^{10}}\)
\(=512.\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}\right)\)
Đặt \(A=1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}\)
\(=>2A=2-1-\frac{1}{2}-\frac{1}{2^2}-....-\frac{1}{2^9}\)
\(=>2A-A=\left(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\right)-\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}\right)\)
\(=>A=2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}\)
\(=>A=2-1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)
\(=>512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}=512.\frac{1}{2^{10}}=\frac{512}{2^{10}}=\frac{1}{2}\)
3 tháng 7 2016
\(=512\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
\(=512\left(1-\frac{1}{2^{10}}\right)=512.\frac{1023}{1024}=\frac{1023}{512}\)
1
A
15 tháng 10 2018
\(\text{M = 512 - 512/2 - .... - 512/2^10
= 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10
= 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2
2M = 2^10 - 2^9 - 2^8 - .... - 1
2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9 + 2^8 + 2^7 +... + 1 + 1/2
M = 2^10 - 2.2^9 + 1/2
M = 2^10 - 2^10 + 1/2}\)
\(\text{ M =}\) \(\frac{1}{2}\)
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
2 tháng 7 2016
\(M=512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}\)
\(M=512-512.\left(\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)
Đặt\(S=\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)
=> \(\frac{1}{2}S=\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{11}}\)
=> \(\frac{1}{2}S-S=-\frac{1}{2}S=\frac{1}{2^{11}}-\frac{1}{2}\)
=> \(S=\left(\frac{1}{2^{11}}-\frac{1}{2}\right):-\frac{1}{2}\)
512-\(\frac{512}{2}\)-\(\frac{512}{2^2}\)-\(\frac{512}{2^3}\)-....-\(\frac{512}{2^{10}}\)
=512-256-\(\frac{2^9}{2^2}\)-\(\frac{2^9}{2^3}\)-\(\frac{2^9}{2^4}\)-\(\frac{2^9}{2^5}\)-\(\frac{2^9}{2^6}\)-\(\frac{2^9}{2^7}\)-\(\frac{2^9}{2^8}\)-\(\frac{2^9}{2^9}\)-\(\frac{2^9}{2^{10}}\)
=512-256-128-64-32-16-8-4-2-\(\frac{1}{2}\)
=\(\frac{3}{2}\)
Đặt \(Q=512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}\)
\(=512-512\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
Đặt A là tên biểu thức trong ngoặc ta cs:
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\)
Thay A vào Q ta được:
\(Q=512-512\left(1-\frac{1}{2^{10}}\right)=512-512+\frac{512}{2^{10}}=\frac{2^9}{2^{10}}=\frac{1}{2}\)