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\(2A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{512}\Rightarrow2A-A=1-\frac{1}{1024}=\frac{1023}{1024}\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)
\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)
\(F=\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^9\div\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
\(F=\frac{\left(\frac{2.5}{5}\right)^7+\left(\frac{9.16}{4.3}\right)^3}{2^7.5^2+2^9}=\frac{2^7+12^3}{2^7.5^2+2^9}=\frac{2^7+2^6.3^3}{2^7.5^2+2^9}=\frac{2^6.\left(2+3^3\right)}{2^7.\left(5^2+2^2\right)}=\frac{2^6.29}{2^7.29}\)
\(F=\frac{1}{2}\)
\(M=512-\frac{512}{2}-\frac{512}{2^2}-\frac{512}{2^3}-...-\frac{512}{2^{10}}\)
\(M=512-\frac{512}{2}-\frac{512}{4}-\frac{512}{8}-...-\frac{512}{1024}\)
\(M=\frac{1024}{2}-\frac{512}{2}-\frac{256}{2}-\frac{128}{2}-...-\frac{1}{2}\)
\(M=\frac{1024}{2}-\left(\frac{512}{2}+\frac{256}{2}+\frac{128}{2}+\frac{64}{2}+...+\frac{1}{2}\right)\)
\(M=\frac{1024}{2}-\frac{1023}{2}\)
\(M=\frac{1}{2}\)
\(M=0,5\)
\(M=512-\frac{512}{2^2}-....-\frac{512}{2^{10}}\)
\(=2^9-\frac{2^9}{2}-.....-\frac{2^9}{10}\)
\(=2^9-2^8-....-\frac{1}{2}\)
\(2M=2^{10}-2^9-....-1\)
\(M=\left(2^{10}-...-1\right)-2^9+2^8+....+1+\frac{1}{2}\)
\(M=2^{10}-2.2^9+\frac{1}{2}\)
\(M=\frac{1}{2}\)
ủa sao ông tự hỏi tự trả lời
Ta có :
M= \(512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}\)=\(512\left(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)
Đặt A= \(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)(1)
=> 2A=\(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^9}\)(2)
Lấy (2)-(1) ta được:
A= \(2+\frac{1}{2^{10}}=\frac{2^{11}+1}{2^{10}}\)
=> M= \(512\cdot\frac{2^{11}+1}{2^{10}}=2^9\cdot\frac{2^{11}+1}{2^{10}}\)= \(\frac{2^{11}+1}{2}=2^{10}+\frac{1}{2}=1024+\frac{1}{2}=\frac{2049}{2}=1024,5\)
Vậy M= \(1024,5\)
bn tự hỏi tự trả lời ak -_-
làm nhầm rồi
Phương An chứ j
Làm lại:
Ta có:
M= \(512-\frac{512}{2}-\frac{512}{2^2}-...-\frac{512}{2^{10}}=512\left(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\right)\)
Đặt A= \(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)(1)
=> 2A=\(2-1-\frac{1}{2}-...-\frac{1}{2^9}\)(2)
Lấy (2)-(1) ta được:
A=\(\frac{1}{2^{10}}\)
=> M= \(512\cdot\frac{1}{2^{10}}=2^9\cdot\frac{1}{2^{10}}=\frac{1}{2}=0,5\)
Vậy M=0,5
M= 0,5 nha