\(A=\dfrac{2008^{2008}+1}{2008^{2009}+1}\)

\(2008\cdot A=\dfrac{2008^{2009}+2008}{2008^{2009}+1}\)

\(=\dfrac{2008^{2009}+1+2007}{2008^{2009}+1}\)

\(=1+\dfrac{2007}{2008^{2009}+1}\)

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

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

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

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

Ta có: \(2008^{2009}+1>2008^{2008}+1\)

\(\Rightarrow\dfrac{1}{2008^{2009}+1}< \dfrac{1}{2008^{2008}+1}\)

\(\Rightarrow\dfrac{2007}{2008^{2009}+1}< \dfrac{2007}{2008^{2008}+1}\)

\(\Rightarrow1+\dfrac{2007}{2008^{2009}+1}< 1+\dfrac{2007}{2008^{2008}+1}\)

hay \(A < B\)

#\(Toru\)

ta có Đặt \(A=\frac{2008^{2008}+1}{2008^{2009}+1}\)

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

Xét A trước ta có

\(2008A=\frac{2008\left(2008^{2008}+1\right)}{2008^{2009}+1}\)\(2008A=\frac{2008^{2009}+2008}{2008^{2009}+1}\)

\(2008A=\frac{2008^{2009}+1+2007}{2008^{2009}+1}\)suy ra \(2008A=1+\frac{2007}{2008^{2009}+1}\)

Xét B ta có 

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

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

VÌ \(1+\frac{2007}{2008^{2009}+1}

Đặt \(a=2008^{2007};\)

\(A=\frac{2008^{2008}+1}{2008^{2009}+1}=\frac{2008a+1}{2008^2.a+1};\text{ }B=\frac{2008^{2007}+1}{2008^{2008}+1}=\frac{a+1}{2008a+1}\)

Quy đồng mẫu ta có:

\(A=\frac{\left(2008a+1\right)\left(2008a+1\right)}{\left(2008^2a+1\right)\left(2008a+1\right)}=\frac{2008^2a^2+2.2008a+1}{\left(2008^2a+1\right)\left(2008a+1\right)}\)

\(B=\frac{\left(a+1\right)\left(2008^2a+1\right)}{\left(2008a+1\right)\left(2008^2a+1\right)}=\frac{2008^2a^2+\left(2008^2+1\right)a+1}{\left(2008a+1\right)\left(2008^2a+1\right)}\)

So sánh ở tử ta thấy \(2.2008

ghi sai đề rồi bạn vế b phải +1 chứ