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PH
1
26 tháng 12 2021
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PH
0
BV
0
MP
1
ND
0
NP
2
12 tháng 2 2020
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 \(A=\frac{512}{2}+\frac{512}{2^2}+\frac{512}{2^3}+...+\frac{512}{2^{10}}\)
\(\Rightarrow2A=512+\frac{512}{2}+\frac{512}{2^2}+...+\frac{512}{2^9}\)
\(\Rightarrow2A-A=512-\frac{512}{2^{10}}\)
\(\Rightarrow A=512-\frac{512}{2^{10}}\)
\(\Rightarrow P=512-A=512-\left(512-\frac{512}{2^{10}}\right)=\frac{512}{2^{10}}=\frac{1}{2}\)
VM
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!
NP
1
DL
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 \(H=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\Leftrightarrow B=512\cdot\left(1-H\right)\)
\(\Leftrightarrow2H=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\\ \Leftrightarrow2H-H=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\\ \Leftrightarrow H=1-\frac{1}{2^{10}}\\ \Leftrightarrow B=512\cdot\left[1-\left(1-\frac{1}{2^{10}}\right)\right]\\ \Leftrightarrow B=512\cdot\frac{1}{2^{10}}\\ \Rightarrow B=2^9\cdot\frac{1}{2^{10}}\\ \Rightarrow B=\frac{1}{2}\)
PQ
0
TB
0
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