Ta có VẾ A

\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)

\(2005\cdot A=\frac{2005\cdot\left(2005^{2005}+1\right)}{2005^{2006}+1}\)

\(2005\cdot A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)

\(2005\cdot A=\frac{2005^{2006}+1+2004}{2005^{2006}+1}\)

\(2005\cdot A=1+\frac{2004}{2005^{2006}+1}\)

Ta lại có Vế B :

\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)

\(2005\cdot B=\frac{2005\cdot\left(2005^{2004}+1\right)}{2005^{2005}+1}\)

\(2005\cdot B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)

\(2005\cdot B=\frac{2005^{2005}+1+2004}{2005^{2005}+1}\)

\(2005\cdot B=1+\frac{2004}{2005^{2005}+1}\)

Nhìn vào trên , suy ra A < B . 

\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2014}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2014}{2005^{2005}+1}=1+\frac{2014}{2005^{2005}+1}\)Ta thấy \(2005^{2006}+1>2005^{2005}+1\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

\(\Rightarrow A< B\)

\(2005A=\frac{2005^{2005}+1}{2005^{2006}+1}=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}}\) \(=\frac{2005^{2006}+2014+1}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005B=\frac{2005^{2004}+1}{2005^{2005}+1}=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}\)\(=\frac{2005^{2005}+2004+1}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Vì \(2005^{2006}+1>2005^{2005}+1\)

Nên \(1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

Hay A < B 

           Vậy A < B 

sửa chỗ \(\frac{2005^{2006}+2014+1}{2005^{2006}+1}\) thành \(\frac{2005^{2006}+2004+1}{2005^{2006}+1}\)nhé 

\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)

\(\Rightarrow2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}\)

\(\Rightarrow2005A=1+\frac{2004}{2005^{2006}+1}\)

\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)

\(\Rightarrow2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}\)

\(\Rightarrow2005B=1+\frac{2004}{2005^{2005}+1}\)

Ta thấy \(\frac{2004}{2005^{2005}+1}>\frac{2004}{2005^{2006}+1}\)

Suy ra \(1+\frac{2004}{2005^{2005}+1}>1+\frac{2004}{2005^{2006}+1}\)

hay 2005B>2005A

Vậy B>A

nhân cả C và D với 2005 rồi tách ra so sánh

Ta có : \(2005C=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005D=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Vì \(\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

=> 2005.C < 2005.D

=> C < D

\(C=\frac{2005^{2005}+1}{2005^{2006}+1}< \frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}=D\)

Vậy \(C< D\)

\(10A=\frac{2005^{2006}+10}{2005^{2006}+1}\)

\(10B=\frac{2005^{2005}+10}{2005^{2005}+1}\)

Rồi bạn so sánh 10A và 10B là ra.

Ai thấy đúng thì ủng hộ nha !!!, sai thì góp ý cho mink nha 

Ta có

A <\(\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)

\(\RightarrowĐPCM\)

\(A=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(B=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Vì  \(2005^{2005}+1< 2005^{2006}+1\)

\(\Rightarrow\frac{2004}{2005^{2005}+1}>\frac{2004}{2005^{2006}+1}\Rightarrow2005B>2005A\Rightarrow B>A\)

 \(2005a=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005b=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

Ta thấy :\(2005^{2006}+1>2005^{2005}+1\)

\(\Rightarrow\frac{2004}{2005^{2006}+1}< \frac{2004}{2005^{2005}+1}\)

\(\Rightarrow1+\frac{2004}{2005^{2006}+1}< 1+\frac{2004}{2005^{2005}+1}\)

\(\Rightarrow2005a< 2005b\)

\(\Rightarrow a< b\)

\(A< \frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}=B\)

Vậy A < B

\(2005A=\frac{2005\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)

\(2005B=\frac{2005\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)

vì 2005²⁰⁰⁶+1>2005²⁰⁰⁵+1

\(\Rightarrow\frac{4}{2005^{2006}+1}< \frac{4}{2005^{2005}+1}\)

\(\Rightarrow1+\frac{4}{2005^{2006}+1}< 1+\frac{4}{2005^{2005}+1}\)

=>A<B

sai đề bài

uk sai ở mẫu của A