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MV
2 tháng 5 2017
1)
\(A=\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\\ 5A=1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\\ 5A-A=\left(1+\dfrac{2}{5}+\dfrac{3}{5^2}+...+\dfrac{500}{5^{49}}\right)-\left(\dfrac{1}{5}+\dfrac{2}{5^2}+\dfrac{3}{5^3}+...+\dfrac{500}{5^{500}}\right)\\ 4A=1-\dfrac{500}{5^{500}}\\ A=\left(1-\dfrac{500}{5^{500}}\right):4\\ A=1:4-\dfrac{500}{5^{500}}:4\\ A=\dfrac{1}{4}-\dfrac{500}{5^{500}\cdot4}< \dfrac{1}{4}< \dfrac{5}{16}\)
Vậy \(A< \dfrac{5}{16}\)
AT
1
S
10 tháng 5 2017
Ta có: \(A=\frac{19^5+2016}{19^5-1}=\frac{19^5-1+2017}{19^5-1}=\frac{19^5-1}{19^5-1}+\frac{2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{19^5-2+2017}{19^5-2}=\frac{19^5-2}{19^5-2}+\frac{2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
Vì \(\frac{2017}{19^5-1}< \frac{2017}{19^5-2}\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\Rightarrow A< B\)
Vậy A < B
LT
1
10 tháng 5 2017
\(A=\frac{19^5+2016}{19^5-1}=\frac{\left(19^5-1\right)+2017}{19^5-1}=1+\frac{2017}{19^5-1}\)
\(B=\frac{19^5+2015}{19^5-2}=\frac{\left(19^5-2\right)+2017}{19^5-2}=1+\frac{2017}{19^5-2}\)
Vì \(19^5-1>19^5-2\) nên \(\frac{2017}{19^5-1}< \frac{2}{19^5-2}\)
\(\Rightarrow1+\frac{2017}{19^5-1}< 1+\frac{2017}{19^5-2}\)
Vậy \(A< B\)
28 tháng 2 2018
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
13 tháng 5 2017
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{200^2}>\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{199.200}=\dfrac{199}{200}>\dfrac{3}{4}\)
MV
19 tháng 1 2018
\(S^2=\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\\ \text{Ta có:}\\ \dfrac{1}{2}< \dfrac{2}{3}\\ \dfrac{3}{4}< \dfrac{4}{5}\\ \dfrac{5}{6}< \dfrac{6}{7}\\ ...\\ \dfrac{199}{200}< \dfrac{200}{201}\\ \Rightarrow S^2< \left(\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{199}{200}\right)\left(\dfrac{2}{3}\cdot\dfrac{4}{5}\cdot\dfrac{6}{7}\cdot...\cdot\dfrac{200}{201}\right)\\ \Leftrightarrow S^2< \dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{199}{200}\cdot\dfrac{200}{201}\\ \Leftrightarrow S^2< \dfrac{1\cdot2\cdot3\cdot...\cdot200}{2\cdot3\cdot4\cdot...\cdot201}\\ \Leftrightarrow S^2< \dfrac{1}{201}< \dfrac{1}{200}\)
Vậy ...
18 tháng 5 2022
a: 17/200>17/314
b: 11/54=22/108<22/37
c: 141/893=3/19
159/901=3/17
mà 3/19<3/17
nên 141/893<159/901
\(\dfrac{1}{2^{500}}=\dfrac{1}{\left(2^5\right)^{100}}=\dfrac{1}{32^{100}}\\ \dfrac{1}{5^{200}}=\dfrac{1}{\left(5^2\right)^{100}}=\dfrac{1}{25^{100}}\)
mà `32^(100)>25^(100)`
nên \(\dfrac{1}{2^{500}}>\dfrac{1}{5^{200}}\)