tính nhanh
A= 1/2+1/4 +1/8+1/16+1/32
tìm A
K
Khách
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DN
1
23 tháng 7 2016
\(A=\frac{1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}}{1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}}\)
Đặt tử số là B, mẫu số là C
\(B=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\)
\(2B=2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\)
\(2B-B=\left(2+1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)\)
\(B=2-\frac{1}{16}\)
\(B=\frac{32}{16}-\frac{1}{16}=\frac{31}{16}\)
\(C=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\)
\(2C=2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\)
\(2C+C=\left(2-1+\frac{1}{2}-\frac{1}{4}+\frac{1}{8}\right)+\left(1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}\right)\)
\(3C=2+\frac{1}{16}\)
\(3C=\frac{32}{16}+\frac{1}{16}\)
\(3C=\frac{33}{16}\)
\(C=\frac{33}{16}:3=\frac{11}{16}\)
=> \(A=\frac{B}{C}=\frac{31}{16}:\frac{11}{16}=\frac{31}{16}.\frac{16}{11}=\frac{31}{11}\)
NC
3
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(\dfrac{4}{2}A=\dfrac{4}{2}\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)\)
\(A=\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+..\left(\dfrac{1}{32}-\dfrac{1}{32}\right)+\left(1-\dfrac{1}{64}\right)\)
\(A=1-\dfrac{1}{64}\)
\(A=\dfrac{63}{64}\)
VD
1
PV
19 tháng 4 2016
Ta có: A =1/2+1/4+1/8+1/16+....+1/256+1/512
=> 2A = 1 + 1/2 + 1/4 + 1/8 + ...+ 1/128 + 1/256
=> 2A - A = (1 + 1/2 + 1/4 + 1/8 + ...+ 1/128 + 1/256 -(1/2+1/4+1/8+1/16+....+1/256+1/512 )
A = 1 - 1/512 = 511/512
8 tháng 4 2021
vì quá dễ nên mình không thể trả lời bạn được nhé!
N
8 tháng 8 2017
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
\(A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}\)
\(2A-A=1-\frac{1}{2^5}\)
\(A=\frac{32}{32}-\frac{1}{32}\)
\(A=\frac{31}{32}\)
7 tháng 7 2016
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{32}-\frac{1}{64}\)
\(=\frac{1}{1}-\frac{1}{64}=\frac{63}{64}\)
7 tháng 7 2016
a) \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{32}\)
\(\Rightarrow2A-A=A=1-\frac{1}{64}=\frac{63}{64}\)
TP
3
1 tháng 8 2016
A = 1/2 + 1/4 + 1/8 + ... + 1/1024
2A = 1 + 1/2 + 1/4 + ... + 1/512
2A - A = (1 + 1/2 + 1/4 + ... + 1/512) - (1/2 + 1/4 + 1/8 + ... + 1/1024)
A = 1 - 1/1024
A = 1023/1024
1 tháng 8 2016
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{1024}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+......+\frac{1}{512}\)
\(\Rightarrow A=2A-A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)
10 tháng 10 2020
TUI ĐANG GẤP CHO TÔI HỎI BÀI NÀY LỚP 2 NHA\\\\
AN CÓ 180 CÁI KẸO.BÌNH CÓ 160. HỎI BÌNH CÓ MẤY CÁI KẸO
10 tháng 10 2020
a) Ta có: \(2.4.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}.\)
\(A=\frac{16}{32}+\frac{8}{32}+\frac{4}{32}+\frac{2}{32}+\frac{1}{32}\)
\(A=\left(\frac{16}{32}+\frac{4}{32}\right)+\left(\frac{8}{32}+\frac{2}{32}\right)+\frac{1}{32}\)
\(A=\frac{20}{32}+\frac{10}{32}+\frac{1}{32}\)
\(A=\frac{31}{32}\)
2A=(1/2+1/4+1/8+1/16+1/32) x2
2A=1+1/2+1/4+1/8+1/16
A =2A-A: Vậy A= 1+1/2+1/4+1/8+1/16
-1/2-1/4-1/8-1/16-1/32
A=32/32-1/32
A=31/32