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

cho j zậy bạn

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

ĐÚNG NÈ

Bạn có nhầm \(\frac{2015}{2}\) thành \(\frac{2015}{1}\) không ?

đề đúng rồi đó bạn

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 ..................................

bạn ơi tách ra thừa số chung rồi làm như bình thường nha 

1, A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}\right)\)

     =4-\(\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}\right)\)

     = 4-\(\left(\frac{1}{1}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{9}\right)\)

    =4-\(\left(1-\frac{1}{9}\right)\)

     = 4-\(\frac{8}{9}\)

      = \(\frac{7}{9}\)

Câu 2 tương tự như câu 1 

A=\(\left(1+1+1+1\right)\)-\(\left(\frac{1}{10}+\frac{1}{40}+...+\frac{1}{154}\right)\)

A= 4                    -\(\left(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{11.14}\right)\)

Bạn tự làm tiếp 

\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right).....\left(1+\frac{1}{9}\right)\left(1+\frac{1}{10}\right)\)

\(=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot.....\cdot\frac{10}{9}\cdot\frac{11}{10}\)

\(=\frac{3.4.5.....10.11}{2.3.4....10}=\frac{11}{2}\)

cảm ơn anh

11/2 nha!

Giúp mình kiếm điểm hỏi đáp nha!