K
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1 tháng 6 2016

A = 2005 x 2005 + 1 / 2005 x 2005 x 2005 - 1

A = 2005  x 2005 + 1 / 2005 x 2005 x 2005 - 1

A = \(\frac{2005+1}{2005x2005-1}\)

B = \(\frac{2005+1}{2005x2005-1}\)

=> A = B

28 tháng 2 2016

A>1>B

=> A>B

25 tháng 6 2015

Xét A trước ta có 

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

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

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

Xét B ta có 

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

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

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

ta có vì 2005A<2005B

từ đó suy ra A<B

 nhớ **** đó

22 tháng 7 2015

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

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

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

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

Vì \(\frac{2004}{2005^{2006}+1}

2 tháng 9 2016

A bé hơn 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 . 

23 tháng 5 2018

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

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

\(\Rightarrow A< B\)

29 tháng 5 2018

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

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

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 

29 tháng 5 2018

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