Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
quy đồngcác phân số lấy mẫu số là 512 .ta có tử số là
256 +128 + 64 +32 +16 +8 +4 +2 +1 =495
A =\(\frac{495}{512}\)
A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
Chúc bạn học giỏi nha!
\(2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A-A=\frac{1}{2}-\frac{1}{512}\)( Cái này dễ tự hiểu nha )
\(A=\frac{255}{512}\)
\(A=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\)
\(A=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(2A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(A=\frac{1}{2}-\frac{1}{2^9}\)
\(A=\frac{2^8}{2^9}-\frac{1}{2^9}\)
\(A=\frac{2^8-1}{2^9}\)
1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 1/2 - 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + ... + 1/256 - 1/512
= 1/2 - 1/512
= 255/512
Gọi \(\frac{1}{4}+\frac{1}{8}+\frac{1}{6}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\) là A
Ta có :
\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(2A=2.\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2A-A=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{11}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(A=\frac{1}{2}-\frac{1}{512}\)
\(A=\frac{255}{512}\)
Vậy ..........
A = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\)+ \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\)+ \(\dfrac{1}{64}\)+ \(\dfrac{1}{128}\)
A\(\times\)2 = 2 + 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
A \(\times\) 2 - A = 2 - \(\dfrac{1}{128}\)
A \(\times\)( 2-1) = \(\dfrac{255}{128}\)
A = \(\dfrac{255}{128}\)
Gọi \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\) là T
\(T=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\)
\(2T=2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\)
\(2T-T=\left(2+1+\dfrac{1}{2}+\dfrac{1}{4}+....+\dfrac{1}{64}\right)-\left(1+\dfrac{1}{2}+....+\dfrac{1}{64}+\dfrac{1}{128}\right)\)
\(T=2+\left(1-1\right)+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+....+\left(\dfrac{1}{64}-\dfrac{1}{64}\right)-\dfrac{1}{128}\)
\(T=2+0+0+...-\dfrac{1}{128}\)
\(T=\dfrac{256}{128}-\dfrac{1}{128}\)
\(T=\dfrac{255}{128}\)
Đặt A = 1/2+1/4+1/8+1/18+1/32+1/64+1/128+1/256
=> 2A = 1+1/2+1/4+1/8+1/18+1/32+1/64+1/128
=> 2A - A = 1 - 1/256
=> A = 255/256 nhé!
Nhận xét :
1/2 = 1 - 1/2 ; 1/4 = 1/2 - 1/4 ; 1/8 = 1/4 - 1/8 ; ..... ; 1/256 = 1/128 - 1/256
=> A = 1 - 1/2 + 1/2 - 1/4 + 1/4 - 1/8 + ..... + 1/128 - 1/256
=> A = 1 - 1/256 = 255/256
đề phải là 1 +1/2 + 1/4 +1/32 + 1/64 + 1/128 +1/256 +/512 +1/1024 moi dug
A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
Chúc bạn học giỏi nha!
gọi biểu thức đó là A
A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/512+1/512
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/256+1/256
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/128+1/128
1/512+A=1/2+1/4+1/8+1/16+1/32+1/64+1/64
1/512+A=1/2+1/4+1/8+1/16+1/32+1/32
1/512+A=1/2+1/4+1/8+1/16+1/16
1/512+A=1/2+1/4+1/8+1/8
1/512+A=1/2+1/4+1/4
1/512+A=1/2+1/2
1/512+A=1
A=1-1/512
A=511/512
chắc 100%
\(ĐặtA=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
\(A=1-\frac{1}{64}=\frac{63}{64}\)
A=a1−r1−rn=41⋅1−211−(21)8=41⋅211−2561=41⋅256255⋅2=512255.
\(A = \frac{255}{512} \approx 0.498046875.\)
Ta có: \(A=\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
=>2×A\(=\frac12+\frac14+\frac18+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
=>2×A-A=\(\frac12+\frac14+\frac18+\cdots+\frac{1}{256}-\frac14-\frac18-\cdots-\frac{1}{512}\)
=>\(A=\frac12-\frac{1}{512}=\frac{256}{512}-\frac{1}{512}=\frac{255}{512}\)
255/512