\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+....+\frac{1}{256}-\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{512}\)

\(=\frac{255}{512}\)

Vậy \(A=\frac{255}{512}\)

A=14 +18 +116 +132 +164 +1128 +1256 +1512 

=12 −14 +14 −18 +....+1256 −1512 

=12 −1512 

=255512 

Vậy A=255512 

Phạm Long Khánh

A = 255/512

Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)

Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)

=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)

Lấy (2) trừ (1) theo vế ta có : 

2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)

\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)

\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)

Xin lỗi bạn Trần thị mai Chi nha mk bấm sai kết quả . Kết quả đúng là : 

\(A=\frac{2^{10}-1}{2^{10}}\)

ta có: 1+1/2+2+1/4+...+9+1/512

        =(1+2+3+4+...+9)+(1/2+1/4+...+1/512)

       =45+(1/2+1/4+...+1/512)

gọi số hạng (1/2+1/4+...+1/512) là a ta được :

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

2a=1+1/2+1/4+1/8+...+1/256

2a-a=(1+1/2+1/4+...+1/256)-(1/2+1/4+...+1/512)

      =1-1/512

      =511/512

vậy kết quả của biểu thức đó là45+511/512

\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)

=\(\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)

=\(\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{4}-\frac{3}{8}+\frac{3}{8}-\frac{3}{16}+\frac{3}{16}-\frac{3}{36}\)

=\(\frac{3}{1}-\frac{3}{36}\)=\(\frac{35}{12}\)

a)=768/512+192/512+48/512+12/512+3/512

=768+192+48+12+3/512

=1023/512 

b)=405/81+135/81+45/81+15/81+5/81

=405+135+45+15+5/81

=595/81

c)=256/192+64/192+16/192+4/192+1/192

=256+64+16+4+1/192

=341/192

cai do ham

Ta có: \(\frac{1}{2}=1-\frac{1}{2}\);  \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\); \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\);  ...; \(\frac{1}{512}=\frac{1}{256}-\frac{1}{512}\); \(\frac{1}{1024}=\frac{1}{512}-\frac{1}{1024}\)

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

            \(=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\)

            \(=1+1-\frac{1}{1024}\)

            \(=2-\frac{1}{1024}=\frac{2047}{1024}\)

bằng 2047/1024

\(\frac{2047}{1024}\)

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-....+\frac{1}{256}-\frac{1}{512}\)

\(=\frac{1}{2}-\frac{1}{512}\)

\(=\frac{255}{512}\)

Vậy \(A=\frac{255}{512}\)

=1/2-1/4+1/4-1/8+1/8-....+1/156-1/152

=1/2-1/152

=255/512

A=255/512

\(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)

\(A=\frac{2-1}{4}+\frac{2-1}{8}+\frac{2-1}{16}+\frac{2-1}{32}+\frac{2-1}{64}+\frac{2-1}{128}+\frac{2-1}{256}+\frac{2-1}{512}\)

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{256}-\frac{1}{512}\)

\(A=\frac{1}{2}-\frac{1}{512}\)

\(A=\frac{256}{512}-\frac{1}{512}=\frac{255}{512}\)

= \(\frac{17}{8}:\frac{25}{14}-\left(15-\frac{40}{3}\right):\frac{25}{6}\)

= \(\frac{17}{8}.\frac{14}{25}-\left(\frac{45}{3}-\frac{40}{3}\right).\frac{6}{25}\)

= \(\frac{119}{100}-\frac{5}{3}.\frac{6}{25}\)  =  \(\frac{119}{100}-\frac{2}{5}\)

=  \(\frac{119}{100}-\frac{40}{100}=\frac{79}{100}\)

Chúc bạn Hk tốt!!!!!

=\(\frac{79}{100}\)

Nhưng mk cần các cách làm ra bạn ơi