câu trả lời của mink là

        1023/1024

Có : 

A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(\Rightarrow2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)

\(\Rightarrow2A-A=\frac{1}{2}-\frac{1}{256}\)

\(A=\frac{128}{256}-\frac{1}{256}=\frac{127}{256}\)

đề phải là 1 +1/2 + 1/4 +1/32 + 1/64 + 1/128 +1/256 +/512 +1/1024 moi dug 

\(A=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{128}-\dfrac{1}{256}\right)\)

\(A=1-\dfrac{1}{256}\)

\(A=\dfrac{255}{256}\)

 A = 1/2 + 1/4 + 1/8 + ... + 1/1024

2A = 1 + 1/2 + 1/4 + ... + 1/512

2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)

A = 1 - 1/1024

A = 1023/1024

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{512}\)

\(\Rightarrow A=2A-A=1-\frac{1}{1024}\)

\(A=\frac{1023}{1024}\)

A=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)

2 x A=\(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\)

2 x A - A = \(\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right)\)

A=\(1-\frac{1}{1024}\)

A=\(\frac{1023}{1024}\)

1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
= 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
= 1 – 1/512
= 511/512

=0.99609375

cách 1 đó

\(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\)

\(2\times E=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\)

\(2\times E-E=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{128}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{256}\right)\)

\(E=1-\dfrac{1}{256}\)

\(E=\dfrac{256}{256}-\dfrac{1}{256}\)

\(E=\dfrac{255}{256}\)

câu 1 . 125

câu 2 .97

k mình nha bn

=255/256

~hok tốt~

#Trang#

Tính \(S=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)

Dùng sai phân như sau

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

Vậy \(S=1-\frac{1}{256}\)

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)

\(=\frac{128}{256}+\frac{64}{256}+\frac{32}{256}+\frac{16}{256}+\frac{8}{256}+\frac{4}{256}+\frac{2}{256}+\frac{1}{256}\)

\(=\frac{128+64+32+16+8+4+2+1}{256}\)

\(=\frac{255}{256}\)

#Hok tốt