Đặt A=1/2+1/4+1/8+..+1/1024

Ax2=1+1/2+1/4+1/8+..+1/512( Nhân cả 2 vế với 2)

Ax2-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 biểu thức đã cho = 1023/1024

giải tri tiết nha

1/5+45/9+1/2+1/3+1/2+1/9+1/15+1/99= ai trả lời đc đưa số tài khoản mik cho 100k

\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+..........+\dfrac{1}{256}+\dfrac{1}{512}=?\)

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

\(=1-\dfrac{1}{512}\)

\(=\dfrac{511}{512}\)

Vậy \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+.........+\dfrac{1}{256}+\dfrac{1}{512}=\dfrac{511}{512}\)

Bài bạn trên cách trình bày mk ko hiểu lắm! mk làm lại nhé!

Đặt :

\(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...........+\dfrac{1}{256}+\dfrac{1}{512}\)

\(\Leftrightarrow S=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+.........+\dfrac{1}{2^8}+\dfrac{1}{2^9}\)

\(\Leftrightarrow2S=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^9}\right)\)

\(\Leftrightarrow2S=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.........+\dfrac{1}{2^8}\)

\(\Leftrightarrow2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^8}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)\)

\(\Leftrightarrow S=1-\dfrac{1}{2^9}\)

\(\Leftrightarrow S=1-\dfrac{1}{512}=\dfrac{511}{512}\)

Ta có : \(\dfrac{1}{2}\) = 1 - \(\dfrac{1}{2}\) ; \(\dfrac{1}{4}\) = \(\dfrac{1}{2}\) - \(\dfrac{1}{4}\) ; \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) - \(\dfrac{1}{8}\) ;............. ;

\(\dfrac{1}{512}\) = \(\dfrac{1}{256}\) - \(\dfrac{1}{512}\)

Vậy : \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + ................... + \(\dfrac{1}{256}\) + \(\dfrac{1}{512}\)

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

= 1 - \(\dfrac{1}{512}\)

= \(\dfrac{511}{512}\)

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

\(=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2A-A=A=1-\dfrac{1}{2^{10}}\)

đây ko phải lớp 5 đúng ko ?

Lời giải:

Sửa lại đề:

\(A=1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-...-\frac{1}{512}\)

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

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

Trừ theo vế:

\(A=2A-A=\frac{1}{2^9}< 0,002\) (đpcm)

\(=\left(2+4+6+...+98\right)\left(6-6\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{512}\right)\)

=0

= ( 2+4+6+...+98 ) ( 6- 6) ( 1/2+1/4 + .......+ 1/ 512 ) 

= 0

Chúc bạn học tốt 

= (2 + 4 + 6 + ... + 98) x ( 6 - 6 ) x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))

= (2 + 4 + 6 + ... + 98) x      0      x ( \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) +...+ \(\dfrac{1}{512}\))

=  0

*****Vì số nào nhân với 0 cũng bằng 0

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

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

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