So sánh : M = 2009^2009 + 1 / 2009^2010 + 1 và N = 2009^2010 - 2 / 2009^2011 - 2
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AT
4
NT
1
27 tháng 1 2021
Ta có :
\(B=\dfrac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Leftrightarrow B< \dfrac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\dfrac{2009^{2010}+2009}{2009^{2011}+2009}=\dfrac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\dfrac{2009^{2009}+1}{2009^{2010}+1}=A\)
\(\Leftrightarrow A>B\)
NT
4
HH
0
LV
0
T0
2
9 tháng 1 2016
\(M=\frac{2009^{2009}+1}{2009^{2010}+1}=\frac{2009^{2009}+1}{2009.2009^{2009}+1}=\frac{1}{2009}\)
\(N=\frac{2009^{2010}-2}{2009^{2011}-2}=\frac{2009^{2010}-2}{2009.2009^{2010}-2}=\frac{1}{2009}\)
Vậy N=M
HN
12 tháng 1 2019
\(B=\frac{2009^{2010}-2}{2009^{2011}-2}< 1\)
\(\Rightarrow B=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}\)\(=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)
Suy ra : \(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2009}+1}{2009^{2010}+1}\) hay \(B< A\)
Vậy \(A>B\)
Ta có :
\(N=\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}\)
\(=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009.\left(2009^{2009}+1\right)}{2009.\left(2009^{2010}+1\right)}\)
\(=\frac{2009^{2009}+1}{2009^{2010}+1}=M\)
Vậy \(M>N\)
Ta có: \(B< 1\)
\(\Rightarrow B< \frac{2009^{2010}-2+3}{2009^{2011}-2+3}=\frac{2009^{2010}+1}{2009^{2011}+1}\left(1\right)\)
Mà \(\frac{2009^{2010}+1}{2009^{2011}+1}< 1\)
\(\Rightarrow\frac{2009^{2010}+1}{2009^{2011}+1}< \frac{2009^{2010}+1+2008}{2009^{2011}+1+2008}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}=A\left(2\right)\)
Từ (1) và (2) suy ra A > B