Ta có : 

\(B=\frac{2004+2005}{2005+2006}=\frac{2004}{2005+2006}+\frac{2005}{2005+2006}< \frac{2004}{2005}+\frac{2005}{2006}=A\)

\(\Rightarrow\)\(B< A\) hay \(A>B\)

Vây \(A>B\)

Chúc bạn học tốt ~

thanks b

Câu này chắc chắn bằng vì hay phân số điều y chan mà 

Bạn vào đay học tham khảo nhé, chắn chắn học xong sẽ biết làm!^^

[Toán nâng cao 6 -7] So sánh lũy thừa ( Tiết 2 ) - YouTube

[Toán nâng cao 6] Dãy phân số viết theo quy luật (Tiết 1 ...

Giải:

Giải theo cách Tổng Hiệu:

Do cOb là góc lớn hơn nên có số đo là:

(150 + 20) : 2 = 85 độ

Số góc aOc là:

150 – 85 = 65 độ

https://www.youtube.com/watch?v=9McmkiUwe-M

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

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

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

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

=> A < B

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

\(\Rightarrow A< B\)

Ta có: \(A=\frac{10^{2004}+1}{10^{2005}+1}\)

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

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

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

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

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

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

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

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

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

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 

\(\frac{2004}{2005}>\frac{2004}{2005+2006}\)

\(\frac{2005}{2006}>\frac{2005}{2005+2006}\)

->\(\frac{2004}{2005}+\frac{2005}{2006}>\frac{2004+2005}{2005+2006}\)

-> A >B

hai biểu thức trên bằng nhau