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

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

17 tháng 11 2018

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-\(\frac{1}{2}\)

=\(\frac{3}{2}\)

17 tháng 11 2018

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:

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