tick đi mình giải cho,dễ ẹc à.

\(1-A=\frac{10^{2007}-10^{2006}}{10^{2007}+1}=\frac{9.10^{2006}}{10^{2007}+1}=\frac{9.2^{2007}}{10^{2008}+10}\)

\(1-B=\frac{10^{2008}-10^{2007}}{10^{2008}+1}=\frac{9.10^{2007}}{10^{2008}+1}\)

=>1-A< 1-B

=> A > B

Ta có: A=\(\frac{10^{2006}+1}{10^{2007}+1}\)

=>10A=\(\frac{10\left(10^{2006}+1\right)}{10^{2007}+1}=\frac{10^{2007}+10}{10^{2007}+1}=1+\frac{9}{10^{2007}+1}\)             

Ta có: B=\(\frac{10^{2007}+1}{10^{2008}+1}\)

=>10B=\(\frac{10\left(10^{2007}+1\right)}{10^{2008}+1}=\frac{10^{2008}+10}{10^{2008}+1}=1+\frac{9}{10^{2008}+1}\)  

Mà \(\frac{9}{10^{2007}+1}>\frac{9}{10^{2008}+1}\)        (do 10²⁰⁰⁷+1<10²⁰⁰⁸+1)

=>\(1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}\)

=>10A>10B

=>A>B

Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)

=> \(B=\frac{10^{2007}+1}{10^{2008}+1}< \frac{10^{2007}+1+9}{10^{2008}+1+9}\)

=> \(B< \frac{10^{2007}+10}{10^{2008}+10}\)

=> \(B< \frac{10.\left(10^{2006}+1\right)}{10.\left(10^{2007}+1\right)}\)

=> \(B< \frac{10^{2006}+1}{10^{2007}+1}=A\)

a<b bn nha

A=\(\frac{10^{2006}+1}{10^{2007}+1}\)

10.A=\(\frac{10.\left(10^{2006}+10\right)}{10^{2007}+1}\)

=\(1+\frac{9}{10^{2007}+1}\)

B=\(\frac{10^{2007}+1}{10^{2008}+1}\)

\(10.B=\frac{10.\left(10^{2007}+10\right)}{10^{2008}+1}\)

= \(1+\frac{9}{10^{2008}+1}\)

Vì\(1+\frac{9}{10^{2007}+1}>1+\frac{9}{10^{2008}+1}\) nên 10A > 10B \(\Rightarrow A>B\)

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