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ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
tách
\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2B-B=\frac{1}{2}-\frac{1}{1024}\)
thay vào B ta có
\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)
\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)
\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)
\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)
\(\Rightarrow A=\frac{513}{1024}\)
Ta có : \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-.....-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\)
=> \(2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{512}\)
=> \(2A-A=\frac{1}{2}-\frac{1}{1024}\)
Thay A vào ta có : \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
Jenny123 tham khảo nhé
Đặt tổng trên là A, ta có:
\(A.2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(A.2-A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{512}-"\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)
\(\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}"\)
\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}-\frac{1}{128}-\frac{1}{256}-\frac{1}{512}-\frac{1}{1024}\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
P/s: Bn xem lại đề nha
ta có 1/19 x 29 + 1/29x39+.........+1/1999x2009
=1/19 - 1/29 . 1/29 - 1/39 ........ 1/1999-1/2009
=1/2009-1/19
=-1990/38171
=>1/19+-1990/38171
=1/2009
K MK MK K LAI
\(a,1-\left(\dfrac{\dfrac{5}{3}}{8}+x-\dfrac{\dfrac{7}{5}}{24}\right)-\dfrac{\dfrac{16}{2}}{3}=0\\ \Leftrightarrow1-\left(\dfrac{5}{24}+x-\dfrac{7}{120}\right)=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{3}{20}+x=1-\dfrac{8}{3}=-\dfrac{5}{3}\\ \Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{20}=-\dfrac{109}{60}\)
\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-.............-\frac{1}{1024}\)
=> 2S = \(2x\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..........-\frac{1}{1024}\right)\)
2S = \(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-..........-\frac{1}{512}\)
2S - S = \(\left(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-........-\frac{1}{512}\right)\)- \(\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-........-\frac{1}{1024}\right)\)
=> S = \(1+\frac{1}{1024}=\frac{1024}{1024}+\frac{1}{1024}=\frac{1025}{1024}\)
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