Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}+\dfrac{1}{128}\\ =\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{64}-\dfrac{1}{128}\\ =\dfrac{1}{2}-\dfrac{1}{128}\\ =\dfrac{63}{128}\)
\(7m^28dm^2=7,08m^2\)
C = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2\(\times\)C = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
2 \(\times\) C - C = 1 - \(\dfrac{1}{128}\)
C = \(\dfrac{127}{128}\)
\(\Leftrightarrow\dfrac{15}{16}:x=1-\dfrac{1}{12}=\dfrac{11}{12}\)
\(\Leftrightarrow x=\dfrac{15}{16}:\dfrac{11}{12}=\dfrac{15}{16}\cdot\dfrac{12}{11}=\dfrac{45}{44}\)
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
M = 1/4 + 1/16 + 1/64 + 1/256 + 1/1024
4.M = 1 + 1/4 + 1/16 + 1/64 + 1/256
4M - M = (1 + 1/4 + 1/16 + 1/64 + 1/256 ) - ( 1/4 + 1/16 + 1/64 + 1/256 + 1/1024 )
3M = 1 - 1/1024
3M = 1023/1024
M = 341/1024
M=\(\dfrac{1}{4}\)+\(\dfrac{1}{16}\)+\(\dfrac{1}{64}\)+\(\dfrac{1}{256}\)+\(\dfrac{1}{1024}\)
=\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\)
=>4M=1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)
=>4M-M=3M=(1+\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\))-(\(\dfrac{1}{4}\)+\(\dfrac{1}{4^2}\)+\(\dfrac{1}{4^3}\)+\(\dfrac{1}{4^4}\)+\(\dfrac{1}{4^5}\))=1-\(\dfrac{1}{4^5}\)=\(\dfrac{1023}{1024}\)
=>M=\(\dfrac{1023}{1024}\):3=\(\dfrac{341}{1024}\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
=>\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
=>\(2A-A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{2}-\dfrac{1}{4}-\dfrac{1}{8}-\dfrac{1}{16}-\dfrac{1}{32}-\dfrac{1}{64}\)
=>\(A=1-\dfrac{1}{64}=\dfrac{63}{64}\)
\(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-...-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}\)
\(=\dfrac{63}{64}\)