Ta thấy: \(\left\{{}\begin{matrix}A=\dfrac{2005^{2014}+1}{2005^{2015}+1}< 1\\B=\dfrac{2005^{2015}+1}{2005^{2016}+1}< 1\end{matrix}\right.\)

\(\Rightarrow\) Áp dụng tính chất \(\dfrac{a}{b}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) ta có:

\(\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2015}+1+2004}{2005^{2016}+1+2004}\)

\(=\dfrac{2005^{2015}+2005}{2005^{2016}+2005}=\dfrac{2005\left(2005^{2014}+1\right)}{2005\left(2005^{2015}+1\right)}=\dfrac{2005^{2014}+1}{2005^{2015}+1}\)

\(\Rightarrow\dfrac{2005^{2015}+1}{2005^{2016}+1}< \dfrac{2005^{2014}+1}{2005^{2015}+1}\)

Vậy \(B< A\)

Hay \(A>B\)

haizzz mk nhớ bài này nhìu người hỏi lắm rồi,chịu khó tìm là thấy

bài này dễ

d​ễ​ thì​ bạn​ làm​ đi

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

Ta thấy:A=\(\frac{2005^{2005+1}}{2005^{2006}+1}\)<1
Ta có:A=\(\frac{2005^{2005}+1}{2005^{2006}+1}\)<\(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\)=\(\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=b
Vậy A<B
Chắc chắn 100%

cái này trong đề cương

\(\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)

\(\dfrac{2005.2006-1}{2004.2006}=1-\dfrac{1}{2005.2006}\)

\(Vì\dfrac{1}{2004.2005}>\dfrac{1}{2005.2006}\Rightarrow1-\dfrac{1}{2004.2005}< 1-\dfrac{1}{2005.2006}\Rightarrow\dfrac{2004.2005-1}{2004.2005}< \dfrac{2005.2006-1}{2004.2006}\)

Ta có :

\(N=\dfrac{-7}{10^{2005}}+\dfrac{-15}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-7}{10^{2006}}+\dfrac{-8}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2006}}\)

\(M=\dfrac{-15}{10^{2005}}+\dfrac{-7}{10^{2006}}=\dfrac{-7}{10^{2005}}+\dfrac{-8}{10^{2005}}+\dfrac{-7}{10^{2006}}=-7\left(\dfrac{1}{10^{2005}}+\dfrac{1}{10^{2006}}\right)+\dfrac{-8}{10^{2005}}\)

Lại có :

\(-\dfrac{8}{10^{2006}}>\dfrac{-8}{10^{2005}}\Leftrightarrow M>N\)

Ta có:

A=(1995+10)x2005

=1995x2005+10x2005

B=(10+2005)x1995

=10x1995+2005x1995

Mà 2005x1995=1995x2005 và 20x1995<20x2005 nên suy ra A>B

K cho mình nha

A>1>B

=> A>B