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\(A=\frac{100^{2007}+1}{100^{2008}+1}\Rightarrow100.A=\frac{100^{2008}+100}{100^{2008}+1}=\frac{100^{2008}+1+99}{100^{2008}+1}=1+\frac{99}{100^{2008}+1}\)
\(B=\frac{100^{2006}+1}{100^{2007}+1}\Rightarrow100.B=\frac{100^{2007}+100}{100^{2007}+1}=\frac{100^{2007}+1+99}{100^{2007}+1}=1+\frac{99}{100^{2007}+1}\)
Vì \(\frac{99}{100^{2007}+1}>\frac{99}{100^{2008}+1};1=1\Rightarrow1+\frac{99}{100^{2007}+1}>1+\frac{99}{100^{2008}+1}\)hay \(A>B\)
Vậy \(A>B\)
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Ta có :
\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=2009\)
Ta có:
\(2007A=\dfrac{2007^{2009}+2007}{2007^{2009}+1}=1+\dfrac{2006}{2007^{2009}+1}\)\(2007B=\dfrac{2007^{2010}+10}{2007^{2010}+1}=1+\dfrac{9}{2007^{2010}+1}\)Vì \(\dfrac{2007}{2007^{2009}+1}>\dfrac{2007}{2007^{2010}+1}\)
=>2007A > 2007B
Do đó A>B
Vậy A>B
Ta có : \(B\) = \(\dfrac{2007^{2009}+1}{2007^{2010}+1}\) \(< 1\) \(\Rightarrow\dfrac{2007^{2009}+1}{2007^{2010}+1}< \dfrac{2007^{2009}+1+2006}{2007^{2010}+1+2006}\) \(=\dfrac{2007^{2009}+2007}{2007^{2010}+2007}\)
\(=\dfrac{2007\left(2007^{2008}+1\right)}{2007\left(2007^{2009}+1\right)}\) \(=\dfrac{2007^{2008}+1}{2007^{2009}+1}=A\)
Vậy \(A>B\)
Áp dụng tính chất : \(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) (\(a;b,m\in N\)*)
Ta có :
\(A=\dfrac{100^{2007}+1}{100^{2008}+1}< \dfrac{100^{2007}+1+99}{100^{2008}+1+99}=\dfrac{100^{2007}+100}{100^{2008}+100}=\dfrac{100\left(100^{2006}+1\right)}{100\left(100^{2007}+1\right)}=\dfrac{100^{2006}+1}{100^{2007}+1}=B\)
\(\Rightarrow A< B\)