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

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

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

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

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

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

\(M=0,5\)

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

ủa sao ông tự hỏi tự trả lời

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

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

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

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

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

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

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 = \(\frac{512}{2^{10}}\)

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

\(\Rightarrow\frac{1}{2}M=\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-\frac{512}{2^4}-...-\frac{512}{2^{11}}\)

\(\Rightarrow M-\frac{1}{2}M=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}-\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+\frac{512}{2^4}+...+\frac{512}{2^{11}}\)

\(\Rightarrow\frac{1}{2}M=512-\frac{512}{2}-\frac{512}{2}-\frac{512}{2^{11}}=-\frac{512}{2^{11}}\Rightarrow M=-\frac{512}{2^{11}}.2=-\frac{512}{2^{10}}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Chúc bạn học tốt!