: Chứng tỏ rằng: 1/20 + 1/21 + 1/23 + ….. + 1/38 + 1/39 > 1/2
giúp mik v m sẽ theo dõi mn
K
Khách
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NM
1
10 tháng 8 2019
Ta có:\(\frac{1}{20}+\frac{1}{21}+...+\frac{1}{27}>\frac{1}{27}+\frac{1}{27}+...+\frac{1}{27}=\frac{8}{27}\)
Vậy đpcm
NT
1
22 tháng 1 2019
\(M=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+.....+\frac{1}{37\cdot38}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{37}-\frac{1}{38}\)
\(=\left(\frac{1}{1}+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{37}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{38}\right)\)
\(=\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{38}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+....+\frac{1}{38}\right)\)
\(=\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{38}\)
\(N=\frac{1}{20\cdot38}+\frac{1}{21\cdot37}+...+\frac{1}{38\cdot20}\)
\(\Rightarrow58N=\frac{1}{20}+\frac{1}{38}+\frac{1}{21}+\frac{1}{37}+...+\frac{1}{37}+\frac{1}{20}\)
\(=2\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{38}\right)\)
\(=2A\)
\(\Rightarrow N=\frac{2}{58}M\)
\(\Rightarrow\frac{M}{N}=29\)là số nguyên.
LT
0
BG
2
TH
Thầy Hùng Olm
Manager
VIP
22 tháng 12 2022
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
JS
1
MV
1 tháng 5 2017
Gọi \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{69}+\dfrac{1}{70}\) là \(S\)
Ta nhận thấy:
\(\dfrac{1}{11},\dfrac{1}{12},\dfrac{1}{13},...,\dfrac{1}{19}\)đều lớn hơn \(\dfrac{1}{20}\)
\(\dfrac{1}{21},\dfrac{1}{22},\dfrac{1}{23},...,\dfrac{1}{29}\)đều lớn hơn \(\dfrac{1}{30}\)
\(\dfrac{1}{31},\dfrac{1}{32},\dfrac{1}{33},...,\dfrac{1}{39}\)đều lớn hơn \(\dfrac{1}{40}\)
\(\dfrac{1}{41},\dfrac{1}{42},\dfrac{1}{43},...,\dfrac{1}{49}\)đều lớn hơn \(\dfrac{1}{50}\)
\(\dfrac{1}{51},\dfrac{1}{52},\dfrac{1}{53},...,\dfrac{1}{59}\)đều lớn hơn \(\dfrac{1}{60}\)
\(\dfrac{1}{61},\dfrac{1}{62},\dfrac{1}{63},...,\dfrac{1}{69}\)đều lớn hơn \(\dfrac{1}{70}\)
\(\Rightarrow S< \dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{30}+...+\dfrac{1}{30}+\dfrac{1}{40}+\dfrac{1}{40}+...+\dfrac{1}{40}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}+\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}+\dfrac{1}{70}+\dfrac{1}{70}+...+\dfrac{1}{70}\\ \Leftrightarrow S< \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\\ =\dfrac{223}{140}\)
\(1\dfrac{5}{29}=\dfrac{34}{29}\)
\(\dfrac{223}{140}>\dfrac{210}{140}=\dfrac{3}{2}=\dfrac{87}{58}>\dfrac{34}{29}\)
Vậy \(\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{69}+\dfrac{1}{70}>1+\dfrac{5}{29}\left(đpcm\right)\)
KT
4
KN
4 tháng 5 2019
Ta có :
\(\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}\)
\(=\frac{1}{3}\left(\frac{3}{20.23}+\frac{3}{23.26}+...+\frac{3}{77.80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+...+\frac{1}{77}-\frac{1}{80}\right)\)
\(=\frac{1}{3}\left(\frac{1}{20}-\frac{1}{80}\right)\)
\(=\frac{1}{3}.\frac{3}{80}\left(\frac{3}{80}< 1\right)\)
\(\Leftrightarrow\frac{1}{20.23}+\frac{1}{23.26}+...+\frac{1}{77.80}< \frac{1}{3}\left(đpcm\right)\)
T
4 tháng 5 2019
\(M=\frac{1}{20.23}+\frac{1}{23.26}+\frac{1}{26.29}+...+\frac{1}{77x80}\)
\(M=\frac{1}{20}-\frac{1}{23}+\frac{1}{23}-\frac{1}{26}+\frac{1}{26}-\frac{1}{29}+...+\frac{1}{77}-\frac{1}{80}\)
\(M=\frac{1}{20}-\frac{1}{80}=\frac{3}{80}\)
\(\frac{3}{80}=\frac{3x9}{80x9}=\frac{27}{720};\frac{1}{9}=\frac{1x80}{9x80}=\frac{80}{720}\)
Vì \(\frac{27}{720}< \frac{80}{720}\Rightarrow\frac{3}{80}< \frac{1}{9}\Rightarrow M< \frac{1}{9}\)
#~Will~be~Pens~#
ta có 1/39<1/20;1/21;...;1/38
=> 1/20 + 1/21 + 1/23 + ….. + 1/38 + 1/39 > 1/39+1/39+...1/39 (20 số 1/39 cộng với nhau)
=>1/20 + 1/21 + 1/23 + ….. + 1/38 + 1/39 > 20/39>1/2