Tính nhanh 3/2 + 3/8 + 3/32 + 3/128 + 3/512
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NP
2
NX
24 tháng 8 2023
Đặt tổng trên là: \(A\)
\(A=\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\)
\(\Rightarrow A.4=6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\)
\(\Rightarrow A.4-A=\left(6+\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}\right)-\left(\dfrac{3}{2}+\dfrac{3}{8}+\dfrac{3}{32}+\dfrac{3}{128}+\dfrac{3}{512}\right)\)
\(\Rightarrow A.3=6-\dfrac{3}{512}=\dfrac{3069}{512}\)
\(\Rightarrow A=\dfrac{3069}{512}:3=\dfrac{1023}{512}\)
HN
8
24 tháng 6 2016
\(P+\frac{1}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{4}{512}=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{4}{128}=\)
\(=\frac{3}{2}+\frac{3}{8}+\frac{4}{32}=\frac{3}{2}+\frac{4}{8}=\frac{4}{2}=2\)
\(\Rightarrow P=2-\frac{1}{512}=\frac{1023}{512}\)
24 tháng 6 2016
\(P=\frac{3}{2}+\frac{3}{8}+...+\frac{3}{512}\)
\(=3.\left(\frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\frac{1}{2^7}+\frac{1}{2^9}\right)\)
\(4P=3\left(\frac{1}{2^3}+\frac{1}{2^5}+...+\frac{1}{2^{11}}\right)\)
\(4P-P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)
\(3P=3\left(\frac{1}{2}-\frac{1}{2^{11}}\right)\)
\(P=\frac{1}{2}-\frac{1}{2^{11}}=\frac{2^{10}-1}{2^{11}}\)
TN
6 tháng 7 2016
Đặt A=3/2+3/8+...+3/512
bn tách
3/2=3/2^1
3/8=3/2^3
....
3/512=3/2^9
Rồi nhân nó lên trừ đc bao nhiêu - đi A ban đầu là đc
TT
2
24 tháng 6 2016
P= 3/2 + 3/8 + 3/32 + 3/128 + 3/512
= 768/512 + 192/512 + 48/512 + 12/512 + 3/512
=1023/512
5 tháng 8 2016
Đặt tổng trên = A
\(A=\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(A.4=6+\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}\)
\(A.4-A=\left(6+\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}\right)-\left(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\right)\)
\(A.3=6-\frac{3}{512}=\frac{3069}{512}\)
\(A=\frac{3069}{512}:3=\frac{1023}{512}\)
5 tháng 8 2016
Đặt A = 3/2 + 3/8 + ... + 3/512
bn tách
3/2 = 3/2^1
3/8 = 3/2^3
...
3/512 = 3/2^9
Rồi nhân nó lên trừ được bao nhiêu - đi A ban đầu là đc
k nka
TT
1
20 tháng 2 2020
\(\frac{3}{2}+\frac{3}{4}+\frac{3}{8}+\frac{3}{16}+\frac{3}{32}+\frac{3}{64}+\frac{3}{128}+\frac{3}{256}+\frac{3}{512}+\frac{3}{1024}\)
=\(3.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}\right)\)
=\(3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}+\frac{1}{128}-\frac{1}{256}+\frac{1}{256}-\frac{1}{512}+\frac{1}{512}-\frac{1}{1024}\right)\)
=\(3.\left(1-\frac{1}{1024}\right)=3.\left(\frac{1024}{1024}-\frac{1}{1024}\right)=3.\frac{1023}{1024}=\frac{3069}{1024}\)
Chúc em học tốt
R
2
17 tháng 7 2017
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}\)
\(=\left(\frac{12}{8}+\frac{3}{8}\right)+\left(\frac{12}{128}+\frac{3}{128}\right)+\frac{3}{512}\)
\(=\frac{15}{8}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{240}{128}+\frac{15}{128}+\frac{3}{512}\)
\(=\frac{255}{128}+\frac{3}{512}\)
\(=\frac{1020}{512}+\frac{3}{512}\)
\(=\frac{1023}{512}\)
17 tháng 7 2017
\(\frac{3}{2}+\frac{3}{8}+\frac{3}{32}+\frac{3}{128}+\frac{3}{512}=\frac{3}{1.2}+\frac{3}{2.4}+\frac{3}{4.8}+\frac{3}{8.16}+\frac{3}{16.32}\)
\(=\frac{3}{1}-\frac{3}{2}+\frac{3}{2}-\frac{3}{4}+\frac{3}{8}-\frac{3}{8}+\frac{3}{16}-\frac{3}{16}+\frac{3}{32}\)
\(=3+\frac{3}{32}=\frac{3.32}{32}+\frac{3}{32}=\frac{96+3}{32}=\frac{99}{32}\)
0,856m=?m
\(...=\dfrac{3}{2}x\left(1+\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}+\dfrac{1}{256}\right)\)
\(=\dfrac{3}{2}x\left(\dfrac{256}{256}+\dfrac{64}{256}+\dfrac{16}{256}+\dfrac{4}{256}+\dfrac{1}{256}\right)\)
\(=\dfrac{3}{2}x\dfrac{341}{256}=\dfrac{1023}{512}\)