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

\(M=512-\frac{512}{2}-\frac{512}{4}-\frac{512}{8}-...-\frac{512}{1024}\)

\(M=\frac{1024}{2}-\frac{512}{2}-\frac{256}{2}-\frac{128}{2}-...-\frac{1}{2}\)

\(M=\frac{1024}{2}-\left(\frac{512}{2}+\frac{256}{2}+\frac{128}{2}+\frac{64}{2}+...+\frac{1}{2}\right)\)

\(M=\frac{1024}{2}-\frac{1023}{2}\)

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

\(M=0,5\)

\(M=512-\frac{512}{2^2}-....-\frac{512}{2^{10}}\)
\(=2^9-\frac{2^9}{2}-.....-\frac{2^9}{10}\)
\(=2^9-2^8-....-\frac{1}{2}\)
\(2M=2^{10}-2^9-....-1\)
\(M=\left(2^{10}-...-1\right)-2^9+2^8+....+1+\frac{1}{2}\)
\(M=2^{10}-2.2^9+\frac{1}{2}\)
\(M=\frac{1}{2}\)

ủa sao ông tự hỏi tự trả lời

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

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

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

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

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

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:

\(Q=512-512\left(1-\frac{1}{2^{10}}\right)=512-512+\frac{512}{2^{10}}=\frac{2^9}{2^{10}}=\frac{1}{2}\)

cau la ai vay minh ten la huy lop 4a

Phùng Nguyễn Quốc Huy Liên quan ko bạn êi, ko trả lời thì thuôi, vào giới thiệu lmj?

dung co ca khia tui

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

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

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

\(=>2A=2-1-\frac{1}{2}-\frac{1}{2^2}-....-\frac{1}{2^9}\)

\(=>2A-A=\left(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\right)-\left(1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-....-\frac{1}{2^{10}}\right)\)

\(=>A=2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{10}}\)

\(=>A=2-1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)

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

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

\(=512\left(1-\frac{1}{2^{10}}\right)=512.\frac{1023}{1024}=\frac{1023}{512}\)

ê câu này mình đi học thêm mãi mới giải dc 

22

\(\text{M = 512 - 512/2 - .... - 512/2^10 = 2^9 - 2^9 / 2 - 2^9/2^2 - ...2^9/2^10 = 2^9 - 2^8 - 2^7 - 2^6 -.... - 1/2 2M = 2^10 - 2^9 - 2^8 - .... - 1 2M - M = 2^10 - 2^9 - 2^8 -... -1 - 2^9 + 2^8 + 2^7 +... + 1 + 1/2 M = 2^10 - 2.2^9 + 1/2 M = 2^10 - 2^10 + 1/2}\)
      \(\text{ M =}\) \(\frac{1}{2}\)

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

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

NÈ :

Bùi Nguyễn Việt Anh

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

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

Đặt\(S=\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)

=> \(\frac{1}{2}S=\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{11}}\)

=> \(\frac{1}{2}S-S=-\frac{1}{2}S=\frac{1}{2^{11}}-\frac{1}{2}\)

=> \(S=\left(\frac{1}{2^{11}}-\frac{1}{2}\right):-\frac{1}{2}\)