\(10A=\dfrac{10^{2007}+10}{10^{2007}+1}=\dfrac{10^{2007}+1+9}{10^{2007}+1}=1+\dfrac{9}{10^{2007}+1}\left(1\right)\)\(10B=\dfrac{10^{2008}+10}{10^{2008}+1}=\dfrac{10^{2008}+1+9}{10^{2008}+1}=1+\dfrac{9}{10^{2008}+1}\left(2\right)\)Từ (1) và ( 2 ) suy ra A>B

Cách 2 :

Ta CM BĐT sau :

\(\dfrac{a}{b}< \dfrac{a+m}{b+m}\left(a< b;a;b;m>0\right)\)

Ta có :

\(a< b\\ \Rightarrow am< bm\\ \Rightarrow ab+am< bm+ab\\ \Rightarrow a\left(b+m\right)< b\left(a+m\right)\\ \Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\)

\(\Rightarrow A=\dfrac{10^{2007}+1}{10^{2008}+1}< \dfrac{10^{2007}+1+9}{10^{2008}+1+9}\\ =\dfrac{10\left(10^{2006}+1\right)}{10\left(10^{2007}+1\right)}=\dfrac{10^{2006}+1}{10^{2007}+1}=B\\ \Rightarrow A< B\)

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

\(10A=\dfrac{10^{2007}+10}{10^{2007}+1}=\dfrac{10^{2007}+1+9}{10^{2007}+1}=1+\dfrac{9}{10^{2007}+1}\left(1\right)\)

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

\(10B=\dfrac{10^{2008}+10}{10^{2008}+1}=\dfrac{10^{2008}+1+10}{10^{2008}+1}=1+\dfrac{9}{10^{2008}+1}\left(2\right)\)

Từ (1)và (2)=>A>B

Chúc Bạn học tốt ,có nhiều thành công trong học tập

10A=10*\(\frac{10^{2006}+1}{10^{2007}+1}\)                             10B=10*\(\frac{10^{2007}+1}{10^{2008}+1}\)                           

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

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

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

Nên 10A>10B=>A>B

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

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

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

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

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

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

Rõ ràng B < 1 nên theo B, nếu \(\frac{a}{b}< 1\) thì \(\frac{a+n}{b+n}>\frac{a}{b}\) => B < \(\frac{\left(10^{2007}+1\right)+9}{\left(10^{2008}+1\right)+9}=\frac{10^{2007}+10}{10^{2008}+10}\)

Do đó B < \(\frac{10^{2007}+10}{10^{2008}+10}=\frac{10\left(10^{2006}+1\right)}{10\left(10^{2007}+1\right)}=\frac{10^{2006}+1}{10^{2007}+1}\)

=> A > B

tick đi mình giải cho,dễ ẹc à.

Áp dụng bất đẳng thức :

\(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\left(a;b;m\in N;b\ne0\right)\)

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

\(\Leftrightarrow B=\dfrac{10^{2007}+1}{10^{2008}+1}< \dfrac{10^{2007}+1+9}{10^{2008}+1+9}=\dfrac{10^{2007}+10}{10^{2008}+10}=\dfrac{10\left(10^{2006}+1\right)}{10\left(10^{2007}+1\right)}=\dfrac{10^{2006}+1}{10^{2007}+1}=A\)

\(\Leftrightarrow B< A\)

a<b bn nha

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*)

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

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

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

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