Question

So sánh : C=1999^1999+1/1999^2000+1 và D=1999^1998+1/1999^1999+1

 

 

 

 

 

 

 

 

 

 

Answer 1

\(C=\frac{1999^{1999}+1}{1999^{2000}+1}<\frac{1999^{1999}+1+1998}{1999^{2000}+1+1998}\)

      \(=\frac{1999^{1999}+1999}{1999^{2000}+1999}\)

      \(=\frac{1999.\left(1999^{1998}+1\right)}{1999.\left(1999^{1999}+1\right)}\)

       \(=\frac{1999^{1998}+1}{1999^{1999}+1}\)\(=D\)

        => C<D

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