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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 \(S=-\frac{1}{64}\)
NM
17
22 tháng 7 2015
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) > 1/40 x 10 = 1/4 (gồm 10 số hạng)
Tương tự : (1/41 + 1/42 + ...+ 1/50) > 1/5 ; (1/51 + 1/52+...+1/59+1/60) > 1/6
S > 1/4 + 1/5 + 1/6.
Trong khi đó (1/4 + 1/5 + 1/6) > 3/5
=>S > 3/5 (1)
S = (1/31+1/32+1/33+...+1/40) + (1/41 + 1/42 + ...+ 1/50) + (1/51 + 1/52+...+1/59+1/60)
Mà : (1/31+1/32+1/33+...+1/40) < 1/31 x 10 = 10/30 = 1/3 (gồm 10 số hạng)
=> S < 4/5 (2)
Từ (1) và (2) => 3/5 <S<4/5
S
12 tháng 1 2023
a) (-37) + 14 + 26 + 37
= [(-37) + 37] + (14 + 26)
= 0 + 40 = 40
b) (-24) + 6 + 10 + 24
= [(-24) + 24] + (10 + 6)
= 0 + 16 = 16
c) 15 + 23 + (-25) + (-23)
= [15 + (-25)] + [23 + (-23)]
= (-10) + 0 = -10
d) 60 + 33 + (-50) + (-33)
= [60 + (-50)] + [33 + (-33)]
= 10 + 0 = 10
e) (-16) + (-209) + (-14) + 209
= [(-16) + (-14)] + [(-209) + 209]
= (-30) + 0 = -30
f) \(-3^2+\left(-54\right)\div\left[\left(-2\right)^8+7\right]\times\left(-2\right)^2\\ =\left(-9\right)+\left(-54\right)\div263\times4\\ =\left(-9\right)+\dfrac{-216}{263}=\dfrac{-2583}{263}\)
12 tháng 1 2023
a. \(\left[\left(-37\right)+37\right]+\left(14+16\right)\) = 30
B. \(\left[\left(-24\right)+24\right]+\left(10+6\right)\) = 16
C. \(\left[\left(-23\right)+23\right]+\left(15-23\right)\)= -8
d. \(\left[33-33\right]+\left(60-50\right)\) = 10
e. \(\left(209-209\right)+\left(-16-14\right)\)= -30
18 tháng 6 2017
Ta có :
\(S=\left(\frac{1}{2}+\frac{3}{4}+\frac{7}{8}+\frac{15}{16}+\frac{31}{32}+\frac{63}{64}+\frac{127}{128}\right)-6\)
\(S=\left(\frac{64}{128}+\frac{102}{128}+\frac{112}{128}+\frac{120}{128}+\frac{124}{128}+\frac{126}{128}+\frac{127}{128}\right)-6\)
\(S=\frac{64+102+112+120+124+126+127}{128}-6\)
\(S=\frac{775}{128}-6\)
\(S=\frac{775}{128}-\frac{768}{128}\)
\(S=\frac{7}{128}\)
18 tháng 6 2017
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)
30 tháng 10 2016
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{49}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(\frac{1}{40}.10+\frac{1}{50}.10+\frac{1}{60}.10< S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50.10}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}\)
\(\frac{1}{4}+\frac{1}{5}+\frac{3}{20}< \frac{1}{4}+\frac{1}{5}+\frac{1}{6}< S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{1}{3}+\frac{4}{15}+\frac{1}{5}\)
\(\frac{3}{5}< S< \frac{4}{5}\left(đpcm\right)\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 6 2023
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)
\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)
\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)
\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)
\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)
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
\(S=\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)
\(S=1+\frac{1}{2}+1+\frac{1}{4}+1+\frac{1}{8}+1+\frac{1}{16}+1+\frac{1}{32}+1+\frac{1}{64}-7\)
\(S=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-1\)
\(S+1=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}\)
\(2\left(S+1\right)=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}\)
\(2\left(S+1\right)-\left(S+1\right)=S+1=1-\frac{1}{2^6}=\frac{63}{64}\)
\(S=\frac{63}{64}-1\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
30 tháng 6 2023
\(S=3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\\ =\left(3+3^2+3^3\right)+3^3.\left(3+3^2+3^3\right)+3^6.\left(3+3^2+3^3\right)\\ =39+3^3.39+3^6.39\\ =-39.\left(-1-3^3-3^6\right)⋮\left(-39\right)\)
NM
30 tháng 6 2023
S = 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39
S = ( 3 + 32 + 33 ) +34 + 35 + 36 + 37 + 38 + 39
S = 39 + 34 + 35 + 36 + 37 + 38 + 39
Vì 39 ⋮ -39
<=> S ⋮ -39
S = 1 + 3 + 32 + 33 +... + 32014
3S = 3 + 32 + 33 + 34 + ... + 32015
3S - S = ( 3 + 32 + 33 + 34 + ... + 32015) - (1 + 3 + 32 + 33 +... + 32014)
2S = 32015 - 1
S = \(\dfrac{3^{2015}-1}{2}\)
Mình vẫn không hiểu lắm!