so sánh\(\frac{2009^{2008+1}}{2009^{2009+1}}\) và \(\frac{2009^{2008+5}}{2009^{2009+9}}\)
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B = 20092009 + 1 / 20092010+1 < 20092009+1+2008 / 20092010+1+2008
= 20092009+2009 / 20092010+2009
= 2009(20092008+1) / 2009(20092009+1)
= 20092008+1 / 20092009+1 = A
=> A > B nhé!
Ai k mk mk k lại !!
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Ta có: \(B=\frac{2009^{2009}+1}{2009^{2010}+1}<\frac{2009^{2009}+1+2008}{2009^{2010}+1+2008}\)
\(=\frac{2009^{2009}+2009}{2009^{2010}+2009}\)
\(=\frac{2009.\left(2009^{2008}+1\right)}{2009.\left(2009^{2009}+1\right)}\)
\(=\frac{2009^{2008}+1}{2009^{2009}+1}=A\)
=> B<A
Ai k mik mik k lại. Chúc các bạn thi tốt
Ta có: $B=\frac{2009^{2009}+1}{2009^{2010}+1}<\frac{2009^{2009}+1+2008}{2009^{2010}+1+2008}$B=20092009+120092010+1 <20092009+1+200820092010+1+2008
$=\frac{2009^{2009}+2009}{2009^{2010}+2009}$=20092009+200920092010+2009
$=\frac{2009.\left(2009^{2008}+1\right)}{2009.\left(2009^{2009}+1\right)}$=2009.(20092008+1)2009.(20092009+1)
$=\frac{2009^{2008}+1}{2009^{2009}+1}=A$=20092008+120092009+1 =A
=> B<A
Ai k mik mik k lại. Chúc các bạn thi tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}=\frac{2009-1}{\sqrt{2009}}+\frac{2008+1}{\sqrt{2008}}=\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)\)
Vì \(\frac{1}{\sqrt{2008}}>\frac{1}{\sqrt{2009}}\) nên \(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}>0\)
\(\Rightarrow\sqrt{2009}+\sqrt{2008}+\left(\frac{1}{\sqrt{2008}}-\frac{1}{\sqrt{2009}}\right)>\sqrt{2009}+\sqrt{2008}\)
Hay \(\frac{2008}{\sqrt{2009}}+\frac{2009}{\sqrt{2008}}>\sqrt{2008}+\sqrt{2009}\)
Ta có:
\(\frac{2009^{2008+1}}{2009^{2009+1}}=\frac{2009^{2009}}{2009^{2010}}=\frac{1}{2009}\)
\(\frac{2009^{2008+5}}{2009^{2009+9}}=\frac{2009^{2013}}{2009^{2018}}=\frac{1}{2009^5}\)
=>Đẳng thức trên lớn hơn đẳng thức dứi(vì 2009<2009^5)
Vậy.......