Cho A=1/11+1/12+1/13+...+1/20
Chứng minh a)A>1/2
b)A<1
K
Khách
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VN
2
DN
14 tháng 3 2018
\(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{70}\)
\(A=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{20}\right)+\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{30}\right)\)
\(+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{70}\right)\)
\(\Rightarrow A< \frac{1}{10}\cdot10+\frac{1}{20}\cdot10+\frac{1}{30}\cdot10+...+\frac{1}{60}\cdot10\)
\(A< 1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{6}\)
\(A< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{6}+\left(\frac{1}{4}+\frac{1}{5}\right)\)
\(A< 2+0,45< 2,5\)
DN
14 tháng 3 2018
\(A=\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{70}\)
\(A>\left(\frac{1}{20}+\frac{1}{20}+..+\frac{1}{20}\right)+\left(\frac{1}{30}+...+\frac{1}{30}\right)+...+\left(\frac{1}{70}+\frac{1}{70}+...+\frac{1}{70}\right)\)
\(A>\frac{1}{2}+\frac{1}{3}+..+\frac{1}{7}\)
\(A>\frac{223}{140}>\frac{4}{3}\)
KI
1
NN
0
NP
0
S
29 tháng 4 2017
a, Ta có: \(A=\frac{1}{11}+\frac{1}{12}+...+\frac{1}{50}=\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{30}\right)+\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+....+\frac{1}{30}>\frac{1}{30}+\frac{1}{30}+...+\frac{1}{30}=\frac{20}{30}=\frac{2}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{20}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{2}{3}+\frac{1}{3}=1>\frac{1}{2}\)
Vậy A > 1/2
b, Ta có: \(\frac{1}{50}>\frac{1}{100};\frac{1}{51}>\frac{1}{100};........;\frac{1}{99}>\frac{1}{100}\)
\(\Rightarrow B>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{50}{100}=\frac{1}{2}\)
Vậy B > 1/2
c, Ta có: \(C=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}=\frac{1}{10}+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}\right)\)
Nhận xét: \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow C>\frac{1}{10}+\frac{9}{10}=\frac{10}{10}=1\)
Vậy C > 1
S
0
A=1/11+1/12+1/13+...+1/20
>1/20+1/20+1/20+...+1/20(10 phân số)
=1/20x10=1/2
vậy A=1/2(đpcm)
b) A=1/11+1/12+1/13+...+1/20
< 1/11+1/11+1/11+...+1/11(10 phân số)
=1/11x1=10/11<11/11=1
vậy A<1(đpcm)