Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
1/2 + 1/4 + 1/8 + 1/16+ 1/32 + 1/64 + 1/128
= 64/ 128 + 32/128 + 16/128 +8/128 + 4/128 +2/128 + 1/128
= ( 64 + 32 + 16 + 8 + 4 + 2 + 1 ) /128
= 127/ 128
= 1 - 1/2 + 1/2 - 1/4 + 1/4 - ............ + 1/64 - 1/128
= 1 - 1/128
= 127/128
k nha bn
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{x}=\dfrac{127}{256}\)
Đặt VT là A
\(\Rightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{2}{x}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{x}\right)=\dfrac{127}{256}\)
\(\Leftrightarrow A=1-\dfrac{1}{x}=\dfrac{127}{256}\)
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{129}{256}\)
\(\Rightarrow x=\dfrac{256}{129}\)
a=511/256
b=647/20
c=mình đang suy nghĩ,nhưng nếu bạn k cho mình thì bạn sẽ có câu trả lời
a. 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
= 1 + ( 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 + 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
= 2 - 1/256
= 511/256
Câu b bạn có viết sai đề không vậy?
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(\dfrac{4}{2}A=\dfrac{4}{2}\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+..\left(\dfrac{1}{32}-\dfrac{1}{32}\right)+\left(1-\dfrac{1}{64}\right)\)
\(A=1-\dfrac{1}{64}\)
\(A=\dfrac{63}{64}\)
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+...+\frac{1}{128}=1-\frac{1}{128}=\frac{127}{128}\)
1/2+1/4+1/8+1/16+1/32+1/64
=(1/2+1/4+1/8)+(1/16+1/32+1/64)
=(4/8+2/8+1/8)+(4/64+2/64+1/64)
=7/8+7/64
=56/64+7/64
=63/64
B = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
2 x B = 1 + \(\dfrac{1}{2}\)+ \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\)+ \(\dfrac{1}{32}\)
2 x B - B = 1 - \(\dfrac{1}{64}\)
B = \(\dfrac{63}{64}\)
1/2+1/4+1/8+1/16+1/32+1/64
Ta thấy:
1/2=1/1-1/2
1/4=1/2-1/4
1/8=1/4-1/8....
1/64=1/32-1/64
A= 1/1-1/2+1/2-1/4+1/4-1/8+.....+1/32-1/64
A=1 - 1/ 64
A= 63/64
\(\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}=\frac{63}{64}\)\(\frac{63}{64}\)