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DT
0
H
26 tháng 2 2020
Theo bài ra ta có :
\(A=\frac{2011}{1.2}+\frac{2011}{3.4}+\frac{2011}{4.5}+...+\frac{2011}{1999.2000}\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{1999.2000}\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1999}-\frac{1}{2000}\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{3}+...+\frac{1}{1999}\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\) \(-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2000}\right)\)
\(\Rightarrow\frac{A}{2011}=\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{2000}\right)-\left(\frac{1}{1}+\frac{1}{2}+...+\frac{1}{1000}\right)\)
\(\Rightarrow\frac{A}{2011}=\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\)
\(\Rightarrow A=2011\left(\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\right)\left(1\right)\)
Ta lại có :
\(B=\frac{2012}{1001}+\frac{2012}{1002}+...+\frac{2012}{2000}\)
\(\Rightarrow B=2012\left(\frac{1}{1001}+\frac{1}{1002}+...+\frac{1}{2000}\right)\)\(\left(2\right)\)
Từ (1) và (2) => A < B
Vậy A < B
NT
0
CA
1
22 tháng 7 2019
So sánh:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}\) và \(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}\)
Ta có:
\(A=-\frac{9}{10^{2012}}-\frac{19}{10^{2011}}=-\frac{1}{10^{2011}}\left(\frac{9}{10}+19\right)=-\frac{1}{10^{2011}}.\frac{199}{10}\)
\(B=-\frac{9}{10^{2011}}-\frac{19}{10^{2012}}=-\frac{1}{10^{2011}}\left(9+\frac{19}{10}\right)=-\frac{1}{10^{2011}}.\frac{109}{10}\)
Vì \(\frac{199}{10}>\frac{109}{10}\Rightarrow\frac{1}{10^{2011}}.\frac{199}{10}>\frac{1}{10^{2011}}.\frac{109}{10}\Rightarrow-\frac{1}{10^{2011}}.\frac{199}{10}< -\frac{1}{10^{2011}}.\frac{109}{10}\)
Vậy nên A<B
BD
28 tháng 9 2017
Bài :1
\(Q=\frac{2010+2011+2012}{2011+2012+2013}\)
\(Q=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow P>Q\)
LM
30 tháng 9 2016
N =\(\frac{2010+2011+2012}{2011+2012+2013}\)
\(\Rightarrow N=\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
Do: \(\frac{2010}{2011}>\frac{2010}{2011+2012+2013};\frac{2011}{2012}>\frac{2011}{2011+2012+2013};\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010}{2011+2012+2013}+\frac{2011}{2011+2012+2013}+\frac{2012}{2011+2012+2013}\)
\(\Rightarrow\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}>\frac{2010+2011+2012}{2011+2012+2013}\Leftrightarrow N>M\)
Ta có : \(\frac{2010}{2011}=\frac{2011}{2011}-\frac{1}{2011}=1-\frac{1}{2011}\)
\(\frac{2011}{2012}=\frac{2012}{2012}-\frac{1}{2012}=1-\frac{1}{2012}\)
mà \(\frac{1}{2011}>\frac{1}{2012}\)
\(\Rightarrow1-\frac{1}{2011}< 1-\frac{1}{2012}\)
\(\Rightarrow\frac{2010}{2011}< \frac{2011}{2012}\)