Đặt \(A=-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{2014}\)

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

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

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

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

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

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

Mình nghĩ đây là nâng cao tiểu học 
Đặt S = 1/2 + 1/4 + 1/8 + 1/16 + ... 
==> 2S = 1 + 1/2 + 1/4 + 1/8 + 1/16 + ... 
2S = 1 + S 
==> S = 1 
Đây cũng là kết quả khi tính theo cấp số nhân khi n --> vô cùng

2A=1+1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512

2A-A=1-1/1024

A=1-1/1024

A=1023/1024

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

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

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

\(\Rightarrow A=\frac{1}{2^{10}-1}\)\(=\frac{1}{1023}\)

nha

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

ta có:

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

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

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

ta có:

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

=> 2B = \(2+1+\frac{1}{2}+...+\frac{1}{512}\)

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

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

=> B = \(\frac{2048}{1024}-\frac{1}{1024}=\frac{2047}{1024}\)

Thay B vào A ta có:

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

vậy A = \(\frac{-2047}{1024}\)

20

\(a)\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)\(=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{5}{24}\)

Hok tốt

Yume Nguyễn bạn giải giúp mk phần b đc k

-1-1/2-1/4-1/8......-1/1024

=-(1+1/2+1/4+1/8...+1/1024)

mà ta có 1024=2^10

nên -(1+1/2+1/4+1/8...+1/1024)

=-(2^9+2^8+2^7....+1)/2^10

=-(1023/1024)

=-1,99.........

mình sẽ làm lại bai này cho đúng nha

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

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

\(=-1-\frac{1023}{1024}=\frac{-1024}{1024}-\frac{1023}{1024}=\frac{-2047}{1024}\)

vậy mới đúng nha

Đặt: \(A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}+\dfrac{1}{1024}\)

\(\Rightarrow\dfrac{4}{2}A=\dfrac{4}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{1024}\right)\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\)

\(\Rightarrow2A-A=\left(3+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{512}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{1024}\right)\)

\(\Rightarrow A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(3-1-\dfrac{1}{1024}\right)\)

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

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

 1+(1)/(2)+(1)/(4)+(1)/(8)+...+(1)/(512)+(1)/(1024)

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

A x 2 = 1 - 1/1024 + A

A x 2 - A = 1 - 1/1024

A = 1 - 1/1024 

A = 1023 /1024

ko bít

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

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

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

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

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

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

=> \(2A-A=1-\frac{1}{1024}\)

=> \(A=\frac{1023}{1024}\)

Tui cux k bt nữa tui tính rồi mà tinhs mãi k ra 

Bà bt ln chưa chỉ tôi vs