Tính:
S = 3/2 + 5/4 + 9/8 + 17/16 + 33/32 + 65/64 - 7
K
Khách
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NH
2
24 tháng 7 2016
\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\)\(\frac{65}{64}-7\)
\(=\frac{96}{64}+\frac{80}{64}+\frac{72}{64}+\frac{66}{64}+\frac{65}{64}-7\)
\(=\frac{96+80+72+66+64}{64}-7\)
\(=\frac{378}{64}-\frac{7}{1}\)
\(=\frac{189}{32}-\frac{224}{32}\)
\(=\frac{-35}{32}\)
27 tháng 6 2016
N = 3/2 + 5/4 + 9/8 + 17/16 + 33/32 + 65/64 - 7
N = (1 + 1/2) + (1 + 1/4) + (1 + 1/8) + (1 + 1/16) + (1 + 1/32) + (1 + 1/64) - 7
N = (1 + 1 + 1 + 1 + 1 + 1 ) + (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) - 7
N = 6 - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64) - 7
Đặt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64
2A = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32
2A - A = (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32) - (1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64)
A = 1 - 1/64
N = 6 - (1 - 1/64) - 7
N = 6 - 1 + 1/64 - 7
N = 5 + 1/64 - 7
N = -2 + 1/64
N = -128/64 + 1/64
N = -127/64
PT
30 tháng 11 2018
bạn soyen_Tiểu bàng giải sao dòng 3 đang cộng đến dòng 9 lại chuyển thành trừ vậy
NT
2
3 tháng 7 2016
\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{31}{32}+\frac{65}{64}-7\)
\(=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+.....+\left(1+\frac{1}{64}\right)-7\)
\(=6+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}\right)-7\)
\(=.....................\)
Q
3 tháng 7 2016
\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\) \(7\)
\(=2\frac{3}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(=3\frac{7}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(=4\frac{15}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(=5\frac{31}{32}+\frac{65}{64}-7\)
\(=6\frac{63}{64}-7\)
\(=-\frac{1}{64}\)
Mình có cách như vậy thôi nhé, vì mình mới học lớp 5
DP
21 tháng 7 2017
\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{4}\right)+\left(1+\frac{1}{8}\right)+\left(1+\frac{1}{16}\right)+\left(1+\frac{1}{32}\right)+\left(1+\frac{1}{64}\right)-7\)
\(S=\left(1+1+....+1\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-7\)
\(S=6+\left[\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+....+\left(\frac{1}{32}-\frac{1}{64}\right)\right]-7\)
\(S=6+\left(1-\frac{1}{64}\right)-7\)
\(S=6+\frac{63}{64}-7\)
\(S=\frac{447}{64}-7=-\frac{1}{64}\)
19 tháng 6 2016
\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-1\)
\(S+1=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(2\left(S+1\right)=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(2\left(S+1\right)-\left(S+1\right)=S+1=1-\frac{1}{2^6}=\frac{63}{64}\)
\(S=\frac{63}{64}-1\)
26 tháng 10 2017
A=(\(\dfrac{1}{3}-\dfrac{1}{3}\))\(+\left(\dfrac{3}{5}+\left(\dfrac{-3}{5}\right)\right)+\left(\dfrac{-5}{7}+\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)\)\(+\left(\dfrac{-11}{13}-\dfrac{9}{11}\right)\)
A\(=0+0+0+0+\dfrac{-238}{143}\)
A\(=\dfrac{-238}{143}\)
\(B=\left(1+\dfrac{1}{2}\right)+\left(1+\dfrac{1}{4}\right)+\left(1+\dfrac{1}{8}\right)+\left(1+\dfrac{1}{32}\right)+\left(1+\dfrac{1}{64}\right)-7\)
\(B=\left(1+1+1+1+1\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\right)-7\)
\(B=6+\dfrac{63}{64}-7\)
\(B=-1+\dfrac{63}{64}\)
\(B=\dfrac{-1}{64}\)
NK
2
20 tháng 6 2017
3/2 + 5/4 + 9/8 + 17/6 + 33/32 + 65/2 - 7
=(3/2+5/4+33/32+65/2) + 9/8+17/6 -7
= 1161/32 + 95/24 - 7
= 3191/26
20 tháng 6 2017
3/2 + 5/4 + 9/8 + 17/6 + 33/32 + 65/2 - 7 = \(\frac{3191}{96}\)
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}-1\)
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 \(S=-\frac{1}{64}\)