Tìm x biết
X*(3+6+12+24+48+96+192+384+768+1536)= 3069
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}+\frac{1}{768}+\frac{1}{1536}\)
\(A\times2=\frac{2}{3}+\frac{2}{6}+\frac{2}{12}+\frac{2}{24}+\frac{2}{48}+\frac{2}{96}+\frac{2}{192}+\frac{2}{384}+\frac{2}{768}+\frac{2}{1536}\)
Rút gọn ta được
\(A\times2=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+\frac{1}{48}+\frac{1}{96}+\frac{1}{192}+\frac{1}{384}+\frac{1}{768}\)
\(A\times2-A=\frac{2}{3}+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{768}-\left[\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1536}\right]\)
\(A=\frac{2}{3}+\frac{1}{3}-\frac{1}{3}-\frac{1}{1536}\)
\(A=\frac{2}{3}-\frac{1}{1536}=\frac{341}{512}\)
1/3 + 1/6 + 1/12 + 1/24 + 1/48 + 1/96 + 1/192 + 1/384 = 85/128
Ta có:
\(S=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\\ =\dfrac{2}{3}+\dfrac{2}{2\times3}+\dfrac{2}{2\times6}+\dfrac{2}{2\times12}+\dfrac{2}{2\times24}+\dfrac{2}{2\times48}+\dfrac{2}{2\times96}+\dfrac{2}{2\times192}\\ =\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}\\ \)
\(\dfrac{S}{2}=\dfrac{1}{2}\left(\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\right)\\ =\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}+\dfrac{1}{384}\)
\(S-\dfrac{S}{2}=\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}-\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}+\dfrac{1}{192}+\dfrac{1}{384}\right)\\ =\dfrac{2}{3}-\dfrac{1}{384}=\dfrac{2\times128-1}{384}\\ =\dfrac{85}{128}\\ \Rightarrow S=\dfrac{85}{128}\times2=\dfrac{85}{64}\)
\(A=\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}+\dfrac{2}{384}\)
\(A.2=\dfrac{4}{3}+\dfrac{2}{3}+\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{24}+\dfrac{2}{48}+\dfrac{2}{96}+\dfrac{2}{192}\)
\(A=A.2-A=\dfrac{4}{3}-\dfrac{2}{384}=\dfrac{127}{96}\)
\(a,\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+....+\frac{1}{384}\)
\(\text{Đ}\text{ặt}\)\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{384}\)
\(2A=\frac{1}{3}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{192}\)
\(2A-A=\frac{1}{3}-\frac{1}{384}\)
a đề sai
b)Đặt A=1/4x7 + 1/7x10 + 1/10x13 +...........+ 1/19x22
\(3A=3\left(\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{19.22}\right)\)
\(3A=\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{19}-\frac{1}{22}\)
\(3A=\frac{1}{4}-\frac{1}{22}\)
\(A=\frac{9}{44}:3\)
\(A=\frac{3}{44}\)
\(A=\dfrac{1}{3}\left(1+\dfrac{1}{2}+...+\dfrac{1}{256}+\dfrac{1}{512}\right)\)
Đặt B=1+1/2+...+1/256+1/512
=>2B=2+1+...+1/128+1/256
=>B=2-1/512=1023/512
=>\(A=\dfrac{1}{3}\cdot\dfrac{1023}{512}=\dfrac{341}{512}\)
X*3069=3069
X=3069/3069
X-1
X = 3069/3069
X = 1
k cho minh nha