Đặt tổng trên là A . Ta có:

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

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

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

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

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

2A=1+1/2+1/4+1/8+.........+1/512

2A‐A=﴾1+1/2+1/4+1/8+....+1/512﴿‐﴾1/2+1/4+1/8+.....+1/1024﴿

A=1‐1/1024 =1023/1024

vậy A=1023/1024

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

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

Từ (1) và (2) \(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{4}+...........+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+..........+\frac{1}{1024}\right)\)

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

\(\Rightarrow A=\frac{2047}{1024}\)

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

\(=\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)

=> \(\frac{1}{2}A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\)

=> \(\frac{1}{2}A-A=\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{11}}\right)-\left(\frac{1}{2^0}+\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)

=> \(-\frac{1}{2}A=\frac{1}{2^{11}}-1=\frac{1-2048}{2048}=-\frac{2047}{2048}\)

=> \(A=-\frac{2047}{2048}:\left(-\frac{1}{2}\right)=\frac{2047}{1024}\)

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

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

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

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

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

Lời giải:

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

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

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

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

$\Rightarrow 3x=1023$

$\Rightarrow x=341$

Đặt $A=\dfrac12+\dfrac14+\dfrac18+\dfrac{1}{16}+...+\dfrac{1}{1024}$

$A=\dfrac12+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}}$

$\dfrac12\cdot A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}}$

$A-\dfrac{1}{2}A=(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{10}})-(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+...+\dfrac{1}{2^{11}})$

$\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^{11}}$

$\dfrac{1}{2}A=\dfrac{1}{2}\cdot(1-\dfrac{1}{2^{10}})$

$\Rightarrow A=1-\dfrac{1}{2^{10}}$

Vậy: ...

$Toru$

dễ 

dễ trong ngoặc đơn

\(-1-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)=-1-\frac{1}{1024}=\frac{-1025}{1024}\)

khó hiểu bạn có thể giải chi tiết giúp mk ko

A = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512 + 1/1024

A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512

A x 2 - A = 1 + 1/2 - 1/2+ 1/4 -1/4 + 1/8 -1/8 + 1/16 -1/16 + ... + 1/512 - 1/512 - 1/1024

A = 1 - 1/1024

A = 1023/1024

la 1023/1024 ban nha

A = 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512 + 1/1024

A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... + 1/512

A x 2 - A = 1 + 1/2 - 1/2+ 1/4 -1/4 + 1/8 -1/8 + 1/16 -1/16 + ... + 1/512 - 1/512 - 1/1024

A = 1 - 1/1024

A = 1023/1024

kết quả là 1/1024