\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)

\(A=1-\dfrac{1}{256}\)

\(A=\dfrac{255}{256}\)

giải tri tiết nha

1/5+45/9+1/2+1/3+1/2+1/9+1/15+1/99= ai trả lời đc đưa số tài khoản mik cho 100k

Đặt A=1/2+1/4+1/8+..+1/1024

Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)

Ax2-A=(1+1/2+1/4+1/8+..+1/512)-(1/2+1/4+1/8+..+1/1024)

<=>A=1-1/1024

<=>A=1023/1024

Vậy biểu thức đã cho = 1023/1024

\(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{128}\\ =\dfrac{1}{2}-\dfrac{1}{128}\\ =\dfrac{63}{128}\)

\(7m^28dm^2=7,08m^2\)

c: =7,08

Đặt \(A=\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{128}\)

\(\Rightarrow A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^7}\\ \Rightarrow2A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^6}\\ \Rightarrow2A-A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^6}-\dfrac{1}{2^2}-\dfrac{1}{2^3}-...-\dfrac{1}{2^7}\\ \Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2^7}=\dfrac{2^6-1}{2^7}=\dfrac{63}{128}\)

\(7m^28dm^2=7\dfrac{8}{100}m^2=7,08m^2\)

M = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 
4.M = 1 + 1/4 + 1/16 + 1/64 + 1/256 
4M - M = (1 + 1/4 + 1/16 + 1/64 + 1/256 ) - ( 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 ) 
3M       = 1 - 1/1024 
 3M       = 1023/1024 
  M        = 341/1024

M=\(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{1024}\)

  =\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\)

=>4M=1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)

=>4M-M=3M=(1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\))-(\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\))=1-\(\dfrac{1}{4^5}\)=\(\dfrac{1023}{1024}\)

=>M=\(\dfrac{1023}{1024}\):3=\(\dfrac{341}{1024}\)

`@` `\text {Ans}`

`\downarrow`

\(M=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}\)

\(\Rightarrow4M=1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\)

\(\Rightarrow4M-M=\) \(\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)-\left(\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}+\dfrac{1}{1024}\right)\)

\(=\) \(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256}-\dfrac{1}{1024}\)

\(=1-\dfrac{1}{1024}\)

\(=\dfrac{1024}{1024}-\dfrac{1}{1024}=\dfrac{1023}{1024}\)

`4M - M = 3M`

 \(\Rightarrow3M=\dfrac{1023}{1024}\) 

\(\Rightarrow M=\dfrac{1023}{1024}\div3\)

\(\Rightarrow M=\dfrac{341}{1024}\)

Vậy, `M = `\(\dfrac{341}{1024}\)

     C =           \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)

  2\(\times\)C =   1 +  \(\dfrac{1}{2}\)  + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) 

2 \(\times\) C - C =   1 -  \(\dfrac{1}{128}\)

       C       = \(\dfrac{127}{128}\)

Sửa đề

\(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}\right)}{\dfrac{1}{4}-\dfrac{1}{6}-\dfrac{1}{64}-\dfrac{1}{256}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}\cdot3+\dfrac{5}{8}=\dfrac{3}{2}+\dfrac{5}{8}=\dfrac{17}{8}\)

A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{\dfrac{4}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2.(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13})}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{4^3}-\dfrac{1}{16^2})}{4.(\dfrac{1}{4})-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{4^2}-\dfrac{1}{16^2})}{4-\dfrac{1}{4^3}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(-\dfrac{1}{16^2})}{4.-\dfrac{1}{4^2}}+\dfrac{5}{8}\)

A= \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{256})}{4.\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.(\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}-\dfrac{1}{4^4})}{4.\dfrac{1}{4}-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

=> \(\dfrac{1}{2}.\dfrac{3.\dfrac{1}{4^3}}{4.}+\dfrac{5}{8}\)

Cách này cũng được và gọn hơn

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

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

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)

đây ko phải lớp 5 đúng ko ?