so sánh:
1/31+1/32+...+1/39+1/40 và 1/4
K
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LN
1
KV
24 tháng 3 2022
Ta có: \(\dfrac{1}{4}=\dfrac{10}{40}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{40}\)
Mà \(\dfrac{1}{31}>\dfrac{1}{40}\)
\(\dfrac{1}{32}>\dfrac{1}{40}\)
\(\dfrac{1}{33}>\dfrac{1}{40}\)
\(\dfrac{1}{34}>\dfrac{1}{40}\)
\(\dfrac{1}{35}>\dfrac{1}{40}\)
\(\dfrac{1}{36}>\dfrac{1}{40}\)
\(\dfrac{1}{37}>\dfrac{1}{40}\)
\(\dfrac{1}{38}>\dfrac{1}{40}\)
\(\dfrac{1}{39}>\dfrac{1}{40}\)
\(\Rightarrow\) \(\dfrac{1}{31}+\dfrac{1}{32}+\dfrac{1}{33}+...+\dfrac{1}{39}+\dfrac{1}{40}>\dfrac{10}{40}=\dfrac{1}{4}\)
Vậy \(S>\dfrac{1}{4}\)
DD
0
DK
0
13 tháng 5 2019
\(M=\frac{10^{30}+1}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10\cdot(10^{30}+1)}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+10}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+1+9}{10^{31}+1}\)
\(\Rightarrow10M=1+\frac{9}{10^{31}+1}\)
\(N=\frac{10^{31}+1}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10\cdot(10^{31}+1)}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+10}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+1+9}{10^{32}+1}\)
\(\Rightarrow10N=1+\frac{9}{10^{32}+1}\)
Mà\(1+\frac{9}{10^{31}+1}>1+\frac{9}{10^{32}+1}\)
Nên \(10M>10N\)
Hay \(M>N\)
NN
3
5 tháng 8 2016
ớ chết, mk nhầm, lm lại nha
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{50}\right)+\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{60}\right)\)
\(S< \frac{1}{30}.10+\frac{1}{40}.10+\frac{1}{50}.10\)
\(S< \frac{1}{3}+\frac{1}{4}+\frac{1}{5}< \frac{4}{5}\)
=> \(S< \frac{4}{5}\)
5 tháng 8 2016
\(S=\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}\)
\(S< 30.\frac{1}{60}\)
\(S< \frac{1}{2}< \frac{4}{5}\)
\(S< \frac{4}{5}\)
CH
2
KV
3 tháng 12 2023
\(32^{15}=\left(2^5\right)^{15}=2^{5.15}=2^{75}\)
\(4^{39}=\left(2^2\right)^{39}=2^{2.39}=2^{78}\)
Do \(2^{78}>2^{75}\)
\(\Rightarrow4^{39}>32^{15}\)
\(\Rightarrow1+4+4^2+...+4^{39}>32^{15}\)
\(\Rightarrow3\left(1+4+4^2+...+4^{39}\right)>32^{15}\)
Vậy \(A>B\)
DH
2
XO
2 tháng 9 2020
Ta có :\(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+...+\frac{1}{90}\right)\)
60 số hạng 30 số hạng 30 số hạng
\(>\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}\right)=30.\frac{1}{60}+30.\frac{1}{90}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{5}{6}\)
2 tháng 9 2020
Ta có: \(\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=30.\frac{1}{60}=\frac{1}{2}\)
Lại có: \(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}+...+\frac{1}{90}>\frac{1}{90}+\frac{1}{90}+...+\frac{1}{90}=30.\frac{1}{90}=\frac{1}{3}\)
\(\Rightarrow\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
\(\Rightarrow\frac{1}{31}+\frac{1}{32}+\frac{1}{33}+...+\frac{1}{90}>\frac{5}{6}\) (đpcm)
Ta có :
\(\frac{1}{31}>\frac{1}{40};\frac{1}{32}>\frac{1}{40};...;\frac{1}{40}=\frac{1}{40}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{39}+\frac{1}{40}\)
\(>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}+\frac{1}{40}=\frac{1}{40}.10=\frac{1}{4}\)
\(\frac{1}{31}+\frac{1}{32}+..................+\frac{1}{39}+\frac{1}{40}>\frac{1}{4}\)
KẾT BẠN NHA