Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có :
\(m.A=\frac{m^{2009}+m}{m^{2009}+1}=\frac{m^{2009}+1+\left(m-1\right)}{m^{2009}+1}=1+\frac{m-1}{m^{2009}+1}\)
\(m.B=\frac{m^{2010}+m}{m^{2010}+1}=\frac{m^{2010}+1+\left(m-1\right)}{m^{2010}+1}=1+\frac{m-1}{m^{2010}+1}\)
Vì m2009+1 < m2010+1 => m.A > m.B => A > B
K NHA BẠN
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
\(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.....+\frac{1}{80}\)
\(=\left(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+\frac{1}{44}+.....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+......+\frac{1}{80}\right)\)
\(>\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}+.....+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\right)\)
\(=\frac{1}{3}+\frac{1}{4}\)
\(=\frac{7}{12}\)
\(B=\frac{2008+2009+2010}{2009+2010+2011}=\frac{2008}{2009+2010+2011}+\frac{2009}{2009+2010+2011}+\frac{2010}{2009+2010+2011}\)
\(< \frac{2008}{2009}+\frac{2009}{2010}+\frac{2010}{2011}=A\)
\(M=\frac{2009^{2009}+1}{2009^{2010}+1}=\frac{2009^{2009}+1}{2009.2009^{2009}+1}=\frac{1}{2009}\)
\(N=\frac{2009^{2010}-2}{2009^{2011}-2}=\frac{2009^{2010}-2}{2009.2009^{2010}-2}=\frac{1}{2009}\)
Vậy N=M
