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NN
1
P
Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
2
31 tháng 12 2023
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
AH
Akai Haruma
Giáo viên
30 tháng 12 2023
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
SH
0
J
3
16 tháng 9 2017
Ta có : \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-.....-\frac{1}{1024}\)
\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\)
=> \(2A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{512}\)
=> \(2A-A=\frac{1}{2}-\frac{1}{1024}\)
Thay A vào ta có : \(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.....+\frac{1}{1024}\right)\)
\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)
16 tháng 9 2017
Jenny123 tham khảo nhé
Đặt tổng trên là A, ta có:
\(A.2=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(A.2-A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{512}-"\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\)
\(\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}+\frac{1}{1024}"\)
\(A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}-\frac{1}{64}-\frac{1}{128}-\frac{1}{256}-\frac{1}{512}-\frac{1}{1024}\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
P/s: Bn xem lại đề nha
CP
3
19 tháng 9 2017
Ta có:
\(-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\right)\)
đặt \(A=1+\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)
\(\frac{1}{2}A=\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^{11}}\)
\(A-\frac{1}{2}A=\frac{1}{2}A\Rightarrow A=\frac{1-\frac{1}{2^{11}}}{\frac{1}{2}}=2-\frac{1}{2^{10}}\)
19 tháng 9 2017
\(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(=-1-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}=\frac{1023}{1024}\)
Vậy, \(-1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}=-1-A=-1-\frac{1023}{1024}=-\frac{2047}{1024}\)
PA
13 tháng 9 2016
\(A=\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)
\(2A=\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\)
\(2A-A=\left(\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-..-\frac{1}{512}\right)-\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\right)\)
\(A=\frac{1}{4}+\frac{1}{4}-\frac{1}{2}+\frac{1}{1024}\)
\(A=\frac{1}{1024}\)
13 tháng 9 2016
\(B=\frac{1}{2}-\frac{1}{4}-...-\frac{1}{1024}\)
\(=-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{1024}\right)\)
\(=-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
Đặt \(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}=A\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)
\(2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)
\(A=1-\frac{1}{2^{10}}\).Thay A vào ta đc: \(B=-\left(1-\frac{1}{2^{10}}\right)\)
\(B=-\left(1-\frac{1}{1024}\right)\)
\(B=-\frac{1023}{1024}\)
NV
1
5 tháng 9 2016
-1-1/2-1/4-1/8-1/16-1/32-1/64-1/128-1/256-1/512-1/1024=-1,9990234375
TV
1
21 tháng 9 2016
Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)
A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)
2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)
2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)) \(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))
A=\(1+\frac{1}{2^{10}}\)
A= \(\frac{1025}{1024}\)