Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn kiểm tra lại đề hộ. Nếu có phân số \(\frac{1}{4}\)thì chịu còn không có thì dễ.
2A=1+1/2+1/4+1/8+.........+1/512
2A-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 A=1023/1024
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}=\frac{31}{64}\)
Chúc bạn học tốt
1/2 + 1/4 + 1/8 + 1/16 + 1/32
=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32
=1-1/32
=31/32
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)
\(=\frac{16+8+4+2+1}{32}\)
\(=\frac{31}{32}\)
Vậy.......
A= 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
2A= 2(1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
= 1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
=>A = 2A-A =1+1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 -1/2 - 1/4 - 1/8 - 1/16 - 1/32 - 1/64 - 1/128 - 1/256
=1-1/256
=255/256
đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{256}\)
=> A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^8}\)
=> 2A=\(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^7}\)
=> 2A-A=\(\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^8}\right)\)
=> A=\(1-\frac{1}{2^8}\)
nếu k quy đồng olaf k tính đc bn nhé
\(\text{Đặt A = }\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2\text{A}=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2\text{A}-\text{A}=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)\)
\(\text{A}=1-\frac{1}{32}=\frac{31}{32}\)