Question

giải giúp mình vsâ 

Answer 1

\(A=\frac{100^{2007}+1}{100^{2009}+1}=>100^2A=\frac{100^2\left(100^{2007}+1\right)}{100^{2009}+1}\)\(=>100^2A=\frac{100^{2009}+100^2}{100^{2009}+1}=\frac{100^{2009}+1+9999}{100^{2009}+1}\)

\(=1+\frac{9999}{100^{2009}+1}\left(1\right)\)

\(B=\frac{100^{2005}+1}{100^{2007}+1}=>100^2B=\frac{100^2\left(100^{2005}+1\right)}{100^{2007}+1}\)\(=>100^2B=\frac{100^{2007}+100^2}{100^{2007}+1}=\frac{100^{2007}+1+9999}{100^{2007}+1}\)

\(=1+\frac{9999}{100^{2007}+1}\left(2\right)\)

Từ (1) và (2) ta thầy  100²⁰⁰⁹+1>100²⁰⁰⁷+1

=>\(\frac{9999}{100^{2009}+1}< \frac{9999}{100^{2007}+1}\)

\(=>1+\frac{9999}{100^{2009}+1}< 1+\frac{9999}{100^{2007}+1}\)=>100²A<100²B=>A<B

 

Answer 2

100²A=\(\frac{100^{2009}+100^2}{100^{2009}+1}=1+\frac{100^2-1}{100^{2009}+1}\)

100²B=\(\frac{100^{2007}+100^2}{100^{2007}+1}=1+\frac{100^2-1}{100^{2007}+1}\)

Vì \(\frac{100^2-1}{100^{2009}+1}< \frac{100^2-1}{100^{2007}+1}\) nên: 100²A<100²B

=>A<B

Answer 3

\(A=\frac{100^{2007}+1}{100^{2009}+1}=\frac{100^{2009}+1-100^2}{100^{2009}+1}=1-\frac{100^2}{100^{2009}+1}\)

\(B=\frac{100^{2005}+1}{100^{2007}+1}=\frac{100^{2007}+1-100^2}{100^{2007}+1}=1-\frac{100^2}{100^{2007}+1}\)

=> A>B