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31 tháng 12 2023

\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)

\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)

\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)

\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

AH
Akai Haruma
Giáo viên
30 tháng 12 2023

Lời giải:

$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$

$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$

$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$

$\frac{3x}{1024}=\frac{1023}{1024}$

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

16 tháng 9 2017

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 : \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)

16 tháng 9 2017

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

19 tháng 9 2017

Ta có: 

\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

đặt \(A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

   \(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)

\(A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}\)

19 tháng 9 2017

\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

Đặt  \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)

Vậy, \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}\)

13 tháng 9 2016

\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(2A=\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\)

\(2A-A=\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)

\(A=\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{1024}\)

\(A=\frac{1}{1024}\)

13 tháng 9 2016

\(B=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)

\(=-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)

\(=-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

Đặt \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}=A\)

\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)

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

\(A=1-\frac{1}{2^{10}}\).Thay A vào ta đc: \(B=-\left(1-\frac{1}{2^{10}}\right)\)

\(B=-\left(1-\frac{1}{1024}\right)\)

\(B=-\frac{1023}{1024}\)

5 tháng 9 2016

-1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024=-1,9990234375

21 tháng 9 2016

Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)

A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)

2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)

2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)\(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))

A=\(1+\frac{1}{2^{10}}\)

A= \(\frac{1025}{1024}\)