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22 tháng 7 2018
Q=\(\dfrac{1}{2}+\left(\dfrac{3}{4}+\dfrac{7}{8}\right)+\left(\dfrac{15}{16}+\dfrac{31}{32}\right)+\left(\dfrac{63}{64}+\dfrac{127}{128}\right)-6\)
Q=\(\dfrac{1}{2}+\dfrac{13}{8}+\dfrac{61}{32}+\dfrac{253}{128}\)\(-6\)
Q= \(\dfrac{64}{128}+\dfrac{208}{128}+\dfrac{244}{128}+\dfrac{253}{128}-6\)
Q= \(\dfrac{769}{128}-6\)
Q=\(\dfrac{769}{128}-\dfrac{768}{128}\)
Q= \(\dfrac{1}{128}\)
TL
3
15 tháng 3 2020
S1 = \(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{127}{128}\)
2S1 = 1 + \(\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+\frac{63}{32}+\frac{127}{64}\)
2S1 - S1 = S1 = 1 + (1 + 1 + 1 + 1 + 1 + 1) - \(\frac{127}{128}\)= 6 + \(\frac{1}{128}\)
=> S = S1 - 6 = 6 + \(\frac{1}{128}\)- 6 = \(\frac{1}{128}\)
15 tháng 3 2020
\(S=\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}-6\)
\(S=\frac{1}{2}+\left(\frac{3}{4}+\frac{7}{8}\right)+\left(\frac{15}{16}+\frac{31}{32}\right)+\left(\frac{63}{64}+\frac{127}{128}\right)-6\)
\(S=\frac{1}{2}+\frac{13}{8}+\frac{61}{32}+\frac{253}{128}-6\)
\(S=\frac{64}{128}+\frac{208}{128}+\frac{244}{128}+\frac{253}{128}-6\)
\(S=\frac{769}{128}-6\)
\(S=\frac{769}{128}-\frac{768}{128}\)
\(S=\frac{1}{128}\)
hok tốt!!
10 tháng 8 2020
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-1\)
Ta đặt: \(P=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\)
=> \(2P=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\)
=> \(2P-P=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)\)
=> \(P=1-\frac{1}{64}\)
Mà \(S=P-1\)
=> \(S=1-\frac{1}{64}-1=-\frac{1}{64}\)
Vậy \(S=-\frac{1}{64}\)
HP
1
XO
7 tháng 7 2019
Đặt A =\(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+...+\frac{127}{128}-6\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{8}\right)+...+\left(1-\frac{1}{128}\right)-6\)
= \(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{2^3}\right)+...+\left(1-\frac{1}{2^7}\right)-6\)(7 cặp số)
= \(1-\frac{1}{2}+1-\frac{1}{2^2}+1-\frac{1}{2^3}+...+1-\frac{1}{2^7}-6\)
= \(\left(1+1+1+...+1\right)-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^7}-6\)
= \(1.7-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-6\)
= \(7-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)-6\)
= \(7-6-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
= \(1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\)
=> 2A = \(2-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\)
Lấy 2A - A = \(\left(2-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^6}\right)\right)-\left(1-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\right)\right)\)
A = \(2-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^6}-1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^7}\)
= \(2-1-1+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^7}-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^6}\right)\)
= \(0+\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^3}+...+\frac{1}{2^6}-\frac{1}{2^6}+\frac{1}{2^7}\right)\)
= \(0+\frac{1}{2^7}\)
= \(\frac{1}{2^7}\)
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
Ta có :
\(S=\left(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}\right)-6\)
\(S=\left(\frac{64}{128}+\frac{102}{128}+\frac{112}{128}+\frac{120}{128}+\frac{124}{128}+\frac{126}{128}+\frac{127}{128}\right)-6\)
\(S=\frac{64+102+112+120+124+126+127}{128}-6\)
\(S=\frac{775}{128}-6\)
\(S=\frac{775}{128}-\frac{768}{128}\)
\(S=\frac{7}{128}\)
S=1/2+3/4+7/8+15/16+31/32+63/64+127/128 -6
S= 1-1/2 + 1-1/4 + 1-1/8 + 1-1/16 + 1-1/32 + 1-1/64+ 1-1/128 - 6
S= (1+1+1+1+1+1+1-6)- (1/2+1/4+1/8+1/16 + 1/32+1/64+1/128)
S= 1- 111/128
S= 17/128
(Làm lụi nha bn)