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\(A=\frac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005A=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=\frac{2005^{2006}+1}{2005^{2006}+1}+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(2005B=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=\frac{2005^{2005}+1}{2005^{2005}+1}+\frac{2004}{2005^{2005}+1}\)
Vì \(\frac{2004}{2005^{2006}+1}
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
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=2005^2005+1/2005^2006+1
=>2005A=2005.(2005^2005+1)/2005^2006+1
=>2005A=2005^2006+2005/2005^2006+1
=>2005A=2005^2006+1+2004/2005^2006+1
=>2005A=2005^2006+1/2005^2006+1 + 1/2005^2006+1
=>2005A=1+1/2005^2006+1
Lại có:B=2005^2004+1/2005^2005+1
=>2005B=2005.(2005^2004+1)/2005^2005+1
=>2005B=2005^2005+2005/2005^2005+1
=>2005B=2005^2005+1+2004/2005^2005+1
=>2005B=2005^2005+1/2005^2005+1 + 1/2005^2005+1
=>2005B=1+1/2005^2005+1
Vì 2006>2005
=>2005^2006>2005^2005
=>2005^2006+1>2005^2005+1
=>1/2005^2006+1<1/2005^2005+1
=>1+1/2005^2006+1<1+1/2005^2005+1
=>2005A<2005B
=>A<B
Vậy A<B
Ủng hộ mik nha mọi người !!!
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2005^{2005}+1}{2005^{2006}+1}< 1\)
\(A< \dfrac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\Rightarrow A< \dfrac{2005^{2005}+2005}{2005^{2006}+2005}\Rightarrow A< \dfrac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\Rightarrow A< \dfrac{2005^{2004}+1}{2005^{2005}+1}=B\)
\(A< B\)
Ta có : A = \(\dfrac{2005^{2005}+1}{2005^{2006}+1}\)
\(2005\)A = \(\dfrac{\left(2005^{2005}+1\right).2005}{2005^{2006}+1}\)
\(2005\)\(A\)= \(\dfrac{2005^{2006}+2005}{2005^{2006}+1}\)
\(2005\)\(A\)= \(\dfrac{2005^{2006}+1+2004}{2005^{2006}+1}\)
\(2005A=\dfrac{2005^{2006}+1}{2005^{2006}+1}+\dfrac{2004}{2005^{2006}+1}\)
\(2005A=1+\dfrac{2004}{2005^{2006}+1}\)
Tương tự như vậy với \(B\) ta đc
\(2005B=1+\dfrac{2004}{2005^{2005}+1}\)
Vì \(2005^{2006}+1>2005^{2005}+1\)
\(=>\) \(1+\dfrac{2004}{2005^{2006}+1}\)\(< \)\(1+\dfrac{2004}{2005^{2005}+1}\)
\(=>\)\(2005A< 2005B\)
\(=>\)\(A< B\)
Vậy \(A< B\)