So sánh :
A = \(\dfrac{2006}{2007}\) - \(\dfrac{2007}{2008}\) + \(\dfrac{2008}{2009}\) - \(\dfrac{2009}{2010}\)
B = \(-\dfrac{1}{2006.2007}\) - \(\dfrac{1}{2008.2009}\)
help me ~ mai mik nộp rồi
K
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TH
30 tháng 8 2018
a) Ta có:
\(-\dfrac{24}{35}< -\dfrac{24}{30}< -\dfrac{19}{30}\)
\(\Rightarrow x< y\)
b) Ta có:
\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)
\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)
\(A=-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
Ta lại có:
\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)
=> Dễ dàng thấy A > B
TN
0
PV
0
MC
6 tháng 3 2019
\(A=\frac{2006}{2007}+\frac{2007}{2008}+\frac{2008}{2009}+\frac{2009}{2006}\)
\(A=\left(1-\frac{1}{2007}\right)+\left(1-\frac{1}{2008}\right)+\left(1-\frac{1}{2009}\right)+\left(1+\frac{3}{2006}\right)\)
\(A=1-\frac{1}{2007}+1-\frac{1}{2008}+1-\frac{1}{2009}+1+\frac{3}{2006}\)
\(A=\left(1+1+1+1\right)-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
\(A=4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)\)
Ta có: \(\left\{{}\begin{matrix}\frac{1}{2007}< \frac{1}{2006}\\\frac{1}{2008}< \frac{1}{2006}\\\frac{1}{2009}< \frac{1}{2006}\end{matrix}\right.\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}< \frac{1}{2006}+\frac{1}{2006}+\frac{1}{2006}=\frac{3}{2006}\)
\(\Rightarrow\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}< 0\)
\(\Rightarrow4-\left(\frac{1}{2007}+\frac{1}{2008}+\frac{1}{2009}-\frac{3}{2006}\right)>4\)
hay \(A>4\)
\(\text{Vậy A>4}\)
29 tháng 11 2020
\(B=1+\left(\frac{2007}{2}+1\right)+\left(\frac{2006}{3}+1\right)+...+\left(\frac{1}{2008}+1\right)=2009\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2008}+\frac{1}{2009}\right)\Rightarrow\frac{A}{B}=\frac{1}{2009}\)
Khó quá bạn ơi !!!
Đợi mk nghĩ chút nha.
hjhj
\(A=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)
\(A=\left(1-\dfrac{1}{2007}\right)-\left(1-\dfrac{1}{2008}\right)+\left(1-\dfrac{1}{2009}\right)-\left(1-\dfrac{1}{2010}\right)\)
\(A=1-\dfrac{1}{2007}-1+\dfrac{1}{2008}+1-\dfrac{1}{2009}-1+\dfrac{1}{2010}\)
\(A=\left(1-1\right)+\left(1-1\right)-\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
\(A=\dfrac{1}{2007}+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2010}\)
\(B=-\dfrac{1}{2006.2007}-\dfrac{1}{2008.2009}\)
\(B=-\left(\dfrac{1}{2006}-\dfrac{1}{2007}\right)-\left(\dfrac{1}{2008}-\dfrac{1}{2009}\right)\)
\(B=-\dfrac{1}{2006}+\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\)
\(B=\dfrac{1}{2007}+\dfrac{1}{2009}-\dfrac{1}{2006}+\dfrac{1}{2008}\)
Dễ dàng thấy \(A>B\)