Bạn chỉ cần ghép sao cho số đó tròn chục là đc
Ta có : \(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\)
\(\Rightarrow2A=1+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^6}+\frac{1}{2^7}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^6}+\frac{1}{2^7}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}+\frac{1}{2^8}\right)\)
\(\Rightarrow A=1-\frac{2}{8}=\frac{256}{256}-\frac{1}{256}=\frac{255}{256}\)
Đặt Z = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
Ta có: \(Z=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2Z=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(2Z=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}\)
\(Z=1-\frac{1}{256}\)
\(Z=\frac{255}{256}\)
ủng hộ lên 1010 điểm nha các bạn
1/2+1/4+1/8+1/1+1/32+1/64+1/128+1/256=S
S*2=1+1/2+1/4+1/8+...........1/128
S*2=1+1-1/2+1/2-1/4+......1/128-1/256
S*2=2-1/256
bước còn lại bạn tự làm
Nguyen van tuan buoc con lai the nao
to ko hieu
