\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}+\frac{3}{256}-\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)

= \(\frac{1.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{4}\right)}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{3}{64}}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{3}{4}+\frac{63}{256}\right)}{\frac{3}{4}+\frac{12}{256}}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.3.\left(\frac{1}{4}+\frac{21}{256}\right)}{3.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}+\frac{17}{256}\right)}{\frac{1}{4}+\frac{1}{64}}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(\frac{3.\left(\frac{1}{4}+\frac{1}{64}\right)+3.\frac{17}{256}:\left(\frac{1}{4}+\frac{1}{64}\right)}{1.\left(\frac{1}{4}+\frac{1}{64}\right)}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\left(3+\frac{51}{256}:\frac{17}{64}\right)+\frac{5}{8}\) 

= \(\frac{1}{2}.\left(3+\frac{3}{4}\right)+\frac{5}{8}\)

= \(\frac{1}{2}.\frac{15}{4}+\frac{5}{8}\)

= \(\frac{15}{8}+\frac{5}{8}\)

= \(\frac{5}{2}\)

\(\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{17}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}-\frac{-5}{8}\)

\(=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{17}}{2.\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{17}\right)}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{4}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

\(=\frac{1}{2}.\frac{111}{68}+\frac{5}{8}\)

\(=\frac{49}{34}\)

Xyz 

bạn sai từ dòng thứ 3 rồi

Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)

\(\Rightarrow A=1-\frac{1}{256}\)

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

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 

cảm ơn

giong bai kia thui

nhung bay gio mk phai off rui

6.30 mk se giai cho nha

chuc bn hoc tot

$$$

Số thích hợp để điền vào chỗ chấm là:397/384

k mk nhé!thanks

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

\(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)

\(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)

\(\frac{1}{16}=\frac{1}{8}-\frac{1}{16}\)\(......\)

Vay B=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+......+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\)

\(B=1-\frac{1}{256}\)

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

\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)

=\(1-\frac{1}{256}\)

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

1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

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

= 255/256

đây là toán lớp 6 sao trông khó khó

    \(B=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{64}-\frac{3}{256}}{1-\frac{1}{4}-\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)

=>\(B=\frac{1.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{13}\right)}{3.\left(\frac{1}{3}-\frac{1}{7}-\frac{1}{14}\right)}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{\frac{4}{4}-\frac{4}{16}-\frac{4}{64}-\frac{4}{256}}+\frac{5}{8}\)

=>\(B=\frac{1}{3}.\frac{3.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}{4.\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{64}-\frac{1}{256}\right)}+\frac{5}{8}\)

=>\(B=\frac{1}{3}.\frac{3}{4}+\frac{5}{8}\)

=>\(B=\frac{1}{4}+\frac{5}{8}\)

=>\(B=\frac{2}{8}+\frac{5}{8}\)

=>\(B=\frac{7}{8}\)

l-i-k-e cho mình nhé bạn.

Lê Chí Cường làm đúng công thức rồi nhưng sai từ bước 2 là B=1(1/3-1/7-1/13)/2(1/3-1/7-1/13) chứ không phải là 3(1/3-1/7-1/13) nha nên kết quả cx sai rồi

\(=1-\frac{1}{256}=\frac{255}{256}\)

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

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{156}\)
Ta có:
\(\frac{1}{2}=1-\frac{1}{2}\)  \(;\)   \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)\(;\)   \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)\(;...;\)   \(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{256}{256}-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Vậy \(A=\frac{255}{256}\)
◘_◘ Đúng 100%

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 =\(\frac{495}{512}\)

cho hỏi làm thế nào để nó ra phân số như thế kia zạ

đáp án là 511/512

Theo đề bài ta có :

\(2B=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)

\(\Leftrightarrow2B-B=\left(1+\frac{1}{2}+...+\frac{1}{128}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)\)

\(\Leftrightarrow B=1-\frac{1}{256}\)

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

\(B=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+..+\frac{1}{256}\)

\(\Rightarrow B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^8}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+..+\frac{1}{2^7}\)

\(\Rightarrow2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)\)

\(\Rightarrow B=1-\frac{1}{2^8}\)

p/s: bài lớp 6!

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

\(B=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}+\frac{1}{2^8}\)

\(\Rightarrow2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}+\frac{1}{2^7}\)

\(\Rightarrow2B-B=1-\frac{1}{2^8}\)

\(B=1-\frac{1}{2^8}\)