\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+........+\frac{1}{1280}\)

\(=\frac{1}{5}+\left(\frac{1}{5}-\frac{1}{10}\right)+\left(\frac{1}{10}-\frac{1}{20}\right)+.....+\left(\frac{1}{640}-\frac{1}{1280}\right)\)

\(=\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{20}+......+\frac{1}{640}-\frac{1}{1280}\)

\(=\frac{1}{5}+\frac{1}{5}-\frac{1}{1280}\)( Tối giản các phân số cho nhau )

\(=\frac{2}{5}-\frac{1}{1280}\)

\(=\frac{511}{1280}\)

\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)

\(=\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\right)\cdot5\cdot\frac{1}{5}\)

\(=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...-\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\left(1+1-\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\left(2-\frac{1}{256}\right)\cdot\frac{1}{5}\)

\(=\frac{511}{256}\cdot\frac{1}{5}\)

\(=\frac{511}{1280}\)

ĐÚNG NÈ

mink cần gấp

\(\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{128}=\frac{1}{x-2}\)

\(\Leftrightarrow\frac{1}{10\cdot1}+\frac{1}{10\cdot2}+\frac{1}{10\cdot3}+\frac{1}{10\cdot4}+...+\frac{1}{10\cdot128}=\frac{1}{x-2}\)

\(\Leftrightarrow\frac{1}{10}\cdot\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)=\frac{1}{x-2}\)

Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\)

\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^6}\)

\(2A-A=2-\frac{1}{2^7}\)

Thay vào biểu thức ta có :

\(\frac{1}{10}\cdot\left(2-\frac{1}{2^7}\right)=\frac{1}{x-2}\)

\(\Leftrightarrow\frac{1}{10}\cdot\frac{255}{128}=\frac{1}{x-2}\Leftrightarrow\frac{51}{256}=\frac{1}{x-2}\)

\(\Leftrightarrow51x-102=256\)

\(51x=358\Rightarrow x=\frac{358}{51}\)

Vậy ..................................

mình cho bạn đó bạn đồng ý nhận lời mời kết bạn từ mình nha!!!!

cho j zậy bạn

\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+......+\frac{1}{1280}\)

\(=\frac{1}{5}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\right)\)

\(=\frac{1}{5}\left[1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+...+\left(\frac{1}{64}-\frac{1}{128}\right)+\left(\frac{1}{128}-\frac{1}{256}\right)\right]\)

\(=\frac{1}{5}\left[1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\right]\)

\(=\frac{1}{5}\left(2-\frac{1}{256}\right)\)

\(=\frac{511}{1280}\)

A = 1/5 + 1/10 + 1/20 + 1/40 + ..... + 1/1280

A x 2 = 2/5 - ( 1 /5 + 1/10 + 1/20 + 1/40 + ... + 1/1280 ) - 1/1280

A x 2 = 2/5 - A - 1/1280

A x 2 - A = 2/5 - 1/1280

A  = 2/5 - 1/1280

A = 511/1280

A = 1/5 + 1/10 + 1/20 + 1/40 + ..... + 1/1280

A x 2 = 2/5 - ( 1 /5 + 1/10 + 1/20 + 1/40 + ... + 1/1280 ) - 1/1280

A x 2 = 2/5 - A - 1/1280

A x 2 - A = 2/5 - 1/1280

A  = 2/5 - 1/1280

A = 511/1280

\(=\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{2}+\frac{1}{5}\cdot\frac{1}{2^2}+\frac{1}{5}\cdot\frac{1}{2^3}+...+\frac{1}{5}\cdot\frac{1}{2^8}\)

\(=\frac{1}{5}\cdot\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)=\frac{1}{5}\cdot2\cdot\left(1-\frac{1}{2}\right)\cdot\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

\(=\frac{2}{5}\cdot\left(1-\frac{1}{2^9}\right)=\frac{2\cdot\left(2^9-1\right)}{5\cdot2^9}=\frac{511}{1280}\)

Gọi tổng trên là A

Ta có : \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+...+\frac{1}{2560}\)

\(2A=2\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)

\(2A=\frac{2}{5}+\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)

\(2A-A=\left(\frac{2}{5}+\frac{1}{5}+...+\frac{1}{1280}\right)-\left(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{2560}\right)\)

\(A\left(2-1\right)=\frac{2}{5}-\frac{1}{2560}\)

\(A.1=\frac{1024}{2560}-\frac{1}{2560}\) 

\(A=\frac{1023}{2560}\)

Ta có : A = 1/5 + 1/10 + 1/20 + ... + 1/2560

2A = 2 ( 1/5 + 1/10 + ... + 1/2560 )

2A = 2/5 + 1/5 + 1/10 + .. + 1/2560

2A - A = ( 2/5 + 1/5 + ... + 1/1280 ) - ( 1/5 + 1/10 + ... + 1/2560 )

A =  2 - 1 = 2/5 - 1/2560

A.1 = 1024/2560 - 1/2560

A = 1023 = 2560

đáp số sai

\(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{15}+.....+\frac{1}{1280}\)

\(A=\frac{1}{5}+\frac{1}{5\times2}+\frac{1}{5\times2\times2}+.....+\frac{1}{5\times2\times2\times2\times2\times2\times2\times2\times2}\)

\(A=\frac{1}{5}\times\left(1+\frac{1}{2}+\frac{1}{2\times2}+.....+\frac{1}{2\times2\times2\times2\times2\times2\times2\times2}\right)\)

\(5\times A=1+\frac{1}{2}+\frac{1}{2\times2}+.....+\frac{1}{2\times2\times2\times2\times2\times2\times2\times2}\)

\(10\times A=2+1+\frac{1}{2}+.....+\frac{1}{2\times2\times2\times2\times2\times2\times2}\)

Lấy hiệu :

\(10\times A-5\times A=2-\frac{1}{2\times2\times2\times2\times2\times2\times2\times2}\)

\(10\times A-5\times A=2-\frac{1}{256}\)

\(5\times A=\frac{2\times256-1}{256}\)

\(5\times A=\frac{511}{256}\)

\(A=\frac{511}{256}\div2\)

\(A=\frac{511}{1280}\)

Đặt : \(A=\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+.....+\frac{1}{1280}\)

\(5A=1+\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+.....+\frac{1}{640}\)

\(5A-A=1-\frac{1}{1280}\)

\(4A=\frac{1279}{1280}\)

\(A=\frac{1279}{1280}.\frac{1}{4}=\frac{1279}{320}\)

=\(\frac{1}{1.5}\)+\(\frac{1}{2.5}\)+\(\frac{1}{4.5}\)+...............................................+\(\frac{1}{256.5}\)

ta rut gon con 1-\(\frac{1}{256}\)

=>ket qua la 255/256

luu y : dau . la dau nhan nhe lop 6 moi hoc

\(=\frac{63}{160}=0,39375\)

=\(=\frac{1}{5}+\frac{1}{2.5}+\frac{1}{4.5}+\frac{1}{4.8}+\frac{1}{8.5.2}+\frac{1}{8.5.4}\)

\(=\frac{1+1+1+1+1+1}{5+2.5+4.5+4.8+8.5.2+8.5}\)

\(=\frac{6}{5.8}\)

\(=\frac{6}{40}\)

mk trả lời nhanh nhất tích đúng cho mk nhé

Sao mà mình hỏi bài này từ lâu lắm rồi mà vẫn chưa có bạn nào trả lời nhỉ?

     A)   \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)

2A= \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)

2A-A = \(1-\dfrac{1}{32}\)

A=  \(\dfrac{31}{32}\)

b)\(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) 

Dặt A=

3A= \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A-A=\(1-\dfrac{1}{792}\) 

2A= \(\dfrac{791}{792}\) 

A= \(\dfrac{791}{792}:2=\dfrac{791}{1584}\)