so sánh A = 2011/1.2 +2011/3.4 +...+2011/1999.2000 ; B = 2012/1001 +2012/1002 +...+2012/2000
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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
GB
0
NB
1
15 tháng 4 2017
Đặt \(A=\dfrac{2011}{1.2}+\dfrac{2011}{3.4}+\dfrac{2011}{5.6}+...+\dfrac{2011}{1999.2000}\)
\(\dfrac{A}{2011}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{1999.2000}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=\left(1+...+\dfrac{1}{1999}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\)
\(=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{1000}\right)\)
\(=\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+...+\dfrac{1}{2000}\)
Vậy \(A=2011\left(\dfrac{1}{1001}+\dfrac{1}{1002}+\dfrac{1}{1003}+...+\dfrac{1}{2000}\right)\)
T
18 tháng 3 2017
theo bài ra ta có:
\(A=\dfrac{2011}{1.2}+\dfrac{2011}{3.4}+...+\dfrac{2011}{1999.2000}\)
\(\Rightarrow\dfrac{A}{2011}=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{1999.2000}\)
\(\Rightarrow\dfrac{A}{2011}=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{3}+...+\dfrac{1}{1999}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\)
\(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\) \(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2000}\right)\) \(\Rightarrow\dfrac{A}{2011}=\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{2000}\right)-\left(\dfrac{1}{1}+\dfrac{1}{2}+...+\dfrac{1}{1000}\right)\) \(\Rightarrow\dfrac{A}{2011}=\dfrac{1}{1001}+\dfrac{1}{1002}+...+\dfrac{1}{2000}\)
\(\Rightarrow A=2011\left(\dfrac{1}{1001}+\dfrac{1}{1002}+...+\dfrac{1}{2000}\right)\left(1\right)\)
ta lại có:
\(B=\dfrac{2012}{1001}+\dfrac{2012}{1002}+...+\dfrac{2012}{2000}\\ \Rightarrow B=2012\left(\dfrac{1}{1001}+\dfrac{1}{1002}+...+\dfrac{1}{2000}\right)\left(2\right)\)
Từ 1 và 2 => A < B\
vậy A < B
HG
14 tháng 12 2016
Ta có:
A=-2012/4025=>-2012/4025x2=-4024/4025
B=-1999/3997=>-1999/3997x2=-3998/3997
Ta có: 4024/4025<1<3998/3997
=>4024/4025<3998/3997
=>-4024/4025>-3998/3997
=>-2012/4025>-1999/3997
18 tháng 4 2017
Có: \(B=\dfrac{2011}{1.2}+\dfrac{2011}{2.3}+\dfrac{2011}{3.4}+...+\dfrac{2011}{1999.2000}\)
B= \(2011\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{1999.2000}\right)\)
B = \(2011\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\right)\)
B= \(2011.\left(1-\dfrac{1}{2000}\right)\)
B = \(2011.\dfrac{1999}{2000}=\dfrac{4019989}{2000}\)