Question

A=1999mũ 1999 +1 / 1999mũ1998 +1

và B=1999mũ2000 +1 / 1999mũ1999

Answer 1

SO SÁNH Avà B á hay tính z bn

Answer 2

#)Giải :

Ta có :

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

\(B=\frac{1999^{2000}+1}{1999^{1999}+1}=\frac{1999^{2000}+1999-1998}{1999^{1999}+1}=1999-\frac{1998}{1999^{1999}+1}\)

Vì \(1999^{1998}+1< 1999^{1999}+1\)

\(\Rightarrow\frac{1}{1999^{1998}+1}>\frac{1}{1999^{1999}+1}\Rightarrow1999+\frac{-1}{1999^{1998}+1}< 1999+\frac{-1}{1999^{1999}+1}\Rightarrow A< B\)