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a,Ta co:\(A=\frac{2005^{2005}+1}{2005^{2006}+1}<\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}\)
\(=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\) =B Vay A<B
b,lam tuong tu nhu y 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
cu lay phep tinh nay tru phep tinh kia hk ra thi nt hoi mink
Mình làm câu a) nha!!!
+) \(A=2009^{2010}+2009^{2009}\)
\(=2009^{2009}.\left(2009+1\right)\)
\(=2009^{2009}.2010\)
+) \(B=2010^{2010}=2010^{2009}.2010\)
Vì \(2010^{2009}>2009^{2009}\)nên \(2010^{2009}.2010>2009^{2009}.2010\)hay \(B>A\)
Vậy \(A< B\)
Hok tốt nha^^
ta có : M = \(\left(1-\frac{1}{8}\right)+\left(1-\frac{2}{9}\right)+\left(1-\frac{3}{10}\right)+...+\left(1-\frac{2009}{2016}\right)\)
M = \(\frac{7}{8}+\frac{7}{9}+\frac{7}{10}+...+\frac{7}{2016}\)
M = \(\frac{21}{24}+\frac{21}{27}+\frac{21}{30}+........+\frac{21}{6048}\)
M = \(21\cdot\left(\frac{1}{24}+\frac{1}{27}+\frac{1}{30}+....+\frac{1}{6048}\right)\)
=> \(\frac{M}{N}=21\)
M=102009+2/102009-1=102009-1+3/102009-1=1+3/102009-1
N=102009/102009-3=102009-3+3/102009-3=1+3/102009-3
vì 102009-1>102009-3
=>m<n