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

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

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

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

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

a: \(\Leftrightarrow\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{9}-\dfrac{1}{10}\right)\cdot\left(x-1\right)+\dfrac{1}{10}x-x=-\dfrac{9}{10}\)

\(\Leftrightarrow\dfrac{9}{10}x-\dfrac{9}{10}-\dfrac{9}{10}x=-\dfrac{9}{10}\)

=>-9/10=-9/10(luôn đúng)

b: \(\Leftrightarrow\dfrac{195x+195+130x+195+117x+195+100x+195}{195}=\dfrac{22\cdot39+4\cdot65+6\cdot39+40\cdot5}{195}\)

=>347x+780=1552

=>347x=772

hay x=772/347