ko bít lm bài này ak! dễ mak

bày cho câu b nha!

umk

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

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

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

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

=1-1/256

=255/256

TGV.Quỷ đúng rr đó

#)Giải :

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

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)

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

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

Lời giải 

\(=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+...+\frac{1}{256}-\frac{1}{512}\)

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

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

Sửa đề :

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

Bài làm :

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

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

\(=\frac{1}{4}-\frac{1}{256}=\frac{63}{256}\)

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

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

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

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

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

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

= (64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/256) X x =1

= 126 X x = 1

x = 1 : 126

x= 1/126

nhớ k mình nha !!!

Có : 

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

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

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

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

Tính nhanh (Tính bằng cách thuận tiện) nhé!!

Đặt A = 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256

=> 2A = 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 

=> 2A - A = 1/4 - 1/256

=> A = 64/256 - 1/256

=> A = 63/256

Đặt \(A=\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+...+\frac{1}{256}\)

\(\Rightarrow A=\frac{1}{2^3}+\frac{1}{2^4}+.....+\frac{1}{2^8}\)

\(\Rightarrow2A=\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^7}\)

\(\Rightarrow2A-A=\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-\)\(\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^8}\right)\)

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

=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