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TT
4
VT
1 tháng 11 2016
1/2+1/4+1/8+1/16+1/32+1/64
=2x(1/2+1/4+1/8+1/16+1/321/64)
=1+1/2+1/4+1/8+1/16+1/32
=1+1/2+1/4+1/8+1/16+1/32-1/2-1/4-1/8-1/16-1/32-1/64
=1-1/64
=63/64
1 tháng 11 2016
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
\(=\frac{32}{64}+\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{1}{64}+\frac{1}{64}\)
\(=\frac{62}{64}=\frac{31}{32}\)
NN
Tính A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
A = ....
Giúp mk nha ! đúng mk sẽ tick cho ^_^ !
3
5 tháng 1 2016
A =1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
A = 64/128 + 32/128 + 16/128 + 8/128 + 4/128 + 2/128 + 1/128
A = 217/218 tick đúng nha
5 tháng 1 2016
\(A=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{128}\)
\(\frac{1}{2}A=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{256}\)
\(A-\frac{1}{2}A=\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{128}-\frac{1}{128}\right)+\left(\frac{1}{2}-\frac{1}{256}\right)\)
\(A=\left(\frac{1}{2}-\frac{1}{256}\right)\times2=1-\frac{1}{128}=\frac{127}{128}\)
JK
30 tháng 7 2019
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{256}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}\right)\)
\(A=1-\frac{1}{512}=\frac{511}{512}\)
30 tháng 7 2019
Đặt: \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\)
\(\Rightarrow2A=2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)
\(\Rightarrow2A-A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}\)\(-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{256}+\frac{1}{512}\right)\)
\(\Rightarrow A=1-\frac{1}{512}\)
\(\Rightarrow A=\frac{511}{512}\)
~ rất vui vì giúp đc bn ~
BT
2
10 tháng 8 2018
1/2 + 1/4 + 1/8 +1/16 + 1/32 + 1/64 + 1/128
= 2 . ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 )
= 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 - 1/128 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 ( Rồi giản ước )
= 1
8 tháng 3 2020
= 128/256 + 64/256 + 32/256 + 16/256 + 8/256 + 4/256 + 2/256 + 1/256
= 255/256
12 tháng 7 2017
\(78-\left(x-25\right).3\frac{1}{3}=73\)
\(78-\left(x-25\right)\cdot\frac{10}{3}=73\)
\(\left(x-25\right)\cdot\frac{10}{3}=78-73\)
\(\left(x-25\right)\cdot\frac{10}{3}=5\)
\(x-25=5:\frac{10}{3}\)
\(x-25=1,5\)
\(x=25+1,5=26,5\)
28 tháng 5 2023
a: 4A=4+4^2+...+4^9
=>3A=4^9-1
=>A=(4^9-1)/3
b: 2A=1+1/2+...+1/2^7
=>A=1-1/256=255/256
c: =1-1/5+1/5-1/9+...+1/85-1/89
=1-1/89=88/89
d: =1/3(3/1*4+3/4*7+...+3/304*307)
=1/3(1-1/4+1/4-1/7+...+1/304-1/307)
=1/3*306/307=102/307
e: E=1-1/2+1/2-1/3+...+1/11-1/12
=1-1/12=11/12
g: =2/5(1-1/6+1/6-1/11+...+1/96-1/101)
=2/5*100/101=40/101
23 tháng 7 2019
B)A*2=(1/2+1/4+....+1/256)*2
=1+1/2+1/4+....+1/128)
A*2-A=(1+1/2+1/4+...+1/128)-(1/2+1/4+...+1/256)
=1-1/256
=255/256
XO
23 tháng 7 2019
a) Đặt A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\)
\(\Rightarrow\frac{1}{3}\times A=\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\)
Lấy \(A-\frac{1}{3}\times A\)theo vế ta có :
\(A-\frac{1}{3}\times A=\left(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\right)-\left(\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\right)\)
\(\Rightarrow\frac{2}{3}\times A=\frac{5}{2}-\frac{5}{486}\)
\(\Rightarrow\frac{2}{3}\times A=\frac{605}{243}\)
\(\Rightarrow A=\frac{605}{243}:\frac{2}{3}\)
\(\Rightarrow A=\frac{605}{162}\)
Vậy \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}=\frac{605}{162}\)
b) Đặt B = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
=> \(\frac{1}{2}\times B=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)
Lấy B trừ \(\frac{1}{2}\times B\)theo vế ta có :
\(B-\frac{1}{2}\times B=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...++\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{512}\right)\)
\(\Rightarrow\frac{1}{2}\times B=\frac{1}{2}-\frac{1}{512}\)
\(\Rightarrow\frac{1}{2}\times B=\frac{255}{512}\)
\(\Rightarrow B=\frac{255}{512}:\frac{1}{2}\)
\(\Rightarrow B=\frac{255}{256}\)
Vậy \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}=\frac{255}{256}\)
1/128+2/128+4/128+8/128+16/128+32/128+64/128=127/128
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