C= 1/5+1/10+1/20+1/40+1/80+...........+1280
1\5+1\10+1\20+1\40+….+1\1280
A = \(\dfrac{1}{5}+\dfrac{1}{10}+...+\dfrac{1}{1280}\)
= \(\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
= \(\dfrac{2}{5}-\dfrac{1}{1280}=\dfrac{511}{1280}\)
Giải:
\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...+\dfrac{1}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{20}+\dfrac{1}{20}-\dfrac{1}{40}+...+\dfrac{1}{640}-\dfrac{1}{1280}\)
\(=\dfrac{2}{5}-\dfrac{1}{1280}\)
\(=\dfrac{511}{1280}\)
1/5+1/10+1/20+1/40+......+1/1280
\(\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+.......+\dfrac{1}{1280}\)
\(=\dfrac{1}{5}+\dfrac{1}{5.2}+\dfrac{1}{5.4}+\dfrac{1}{5.8}+.....+\dfrac{1}{5.256}\)
\(=\dfrac{1}{5}+\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+......+\dfrac{1}{2^8}\right)\)
Đặt \(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.......+\dfrac{1}{2^8}\)
\(2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^7}\)
\(2A-A=2-\dfrac{1}{2^3}\)
Thay A vào ta được :
\(\dfrac{1}{5}.A=\dfrac{1}{5}.\left(2-\dfrac{1}{2^8}\right)=\dfrac{511}{1280}\)
1/5 + 1/10 + 1/20 + 1/40 + .... +1/1280
\(\frac{1}{5}+\frac{1}{10}+\frac{1}{20}+\frac{1}{40}+...+\frac{1}{1280}\)
\(=\frac{1}{5}\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)\)
\(=\frac{\frac{1}{5}\left(1-\frac{1}{2^9}\right)}{\left(1-\frac{1}{2}\right)}\)
\(=\frac{2}{5}\left(1-\frac{1}{2^9}\right)\)
\(\frac{1}{5}+\frac{1}{10}+...+\frac{1}{1280}\)
\(=\frac{1}{5}\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
Đặt \(A=1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(2A=2\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(2A=2+1+...+\frac{1}{2^7}\)
\(2A-A=\left(2+1+...+\frac{1}{2^7}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2^8}\right)\)
\(A=2-\frac{1}{2^8}\).Thay A vào đc: \(\frac{1}{5}\cdot\left(2-\frac{1}{256}\right)=\frac{1}{5}\cdot\frac{511}{256}=\frac{511}{1280}\)
tính nhanh:1/5+1/10+1/20+1/40+........+1/1280
1/5 + 1/5 - 1/10 + 1/10 - 1/20 + 1/20 - 1/40 + ... + 1/640 - 1/1280
= 1/5 + 1/5 - 1/1280 = 511/1280
tính nhanh 1/5+1/10+1/20+1/40+...+1/1280
tính nhanh 1/5+1/10+1/20+1/40+...+1/1280
Cho mình hỏi:
1 phần 5 + 1 phần 10 + 1 phần 20 + 1 phần 40 + 1 phần 80 + 1 phần 160 + 1 phần 320 + 1 phần 640 + 1 phần 1280= bao nhiêu ạ
Mong mọi người trả lời sớm mình cần gấp!
A = 1/5 + 1/10 + 1/20 + 1/40 + 1/80 +1/160 + 1/320 + 1/640 + 1/1280
A x 2 = 2/5 + 1/5 + 1/10 + 1/20 + 1/40 + 1/80 +1/160 + 1/320 + 1/640
A x 2 - A = A = 2/5 - 1/1280 = 511/1280
Tính nhanh
1/5+1/10+1/20+1/40+...+1/1280
1/5+1/10+1/20+1/40+.........+1/1280
Giải hộ mình với mình đang vội.