Cho số a bất kì.CMR \(\dfrac{a^{2012}+2012}{\sqrt{a^{2012}+2011}}>2\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(\sqrt{a^{2012}+2011}\le\dfrac{a^{2012}+2011+1}{2}\)
\(\Leftrightarrow\dfrac{a^{2012}+2012}{\sqrt{a^{2012}+2011}}\ge\dfrac{a^{2012}+2012}{\dfrac{a^{2012}+2012}{2}}=2\)
Dấu \("="\Leftrightarrow a^{2012}+2011=1\Leftrightarrow a\in\varnothing\)
Vậy dấu \("="\) ko xảy ra
\(\Rightarrow\dfrac{a^{2012}+2012}{\sqrt{a^{2012}+2011}}>2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(\sqrt{2011}=a;\sqrt{2012}=b\)
Theo đề, ta có: \(A=\dfrac{a^2}{b}+\dfrac{b^2}{a}=\dfrac{a^3+b^3}{ab}\)
B=a+b
\(A-B=\dfrac{a^3+b^3}{ab}-\left(a+b\right)=\dfrac{a^3+b^3-a^2b-ab^2}{ab}\)
\(=\dfrac{\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)}{ab}\)
\(=\dfrac{\left(a+b\right)\left(a-b\right)^2}{ab}>0\)
=>A>B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(B=\dfrac{2011+2012}{2012+2013}=\dfrac{2011}{2012+2013}=\dfrac{2012}{2012+2013}\)
Mà : \(\dfrac{2011}{2012}>\dfrac{2011}{2012+2013}\)
\(\dfrac{2012}{2013}>\dfrac{2012}{2012+2013}\)
\(\Rightarrow \dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)
\(\Rightarrow\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011+2012}{2012+2013}\)
Vậy A > B
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có
A=\(\dfrac{2011+2012}{2012+2013}\)=\(\dfrac{2011}{2012+2013}\)+\(\dfrac{2012}{2012+2013}\)(1)
B=\(\dfrac{2011}{2012}\)+\(\dfrac{2012}{2013}\)(2)
=>A>B
A lớn
B nhỏ
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
em mới lớp 9 nên ko biết anh thông cảm nhé
\(VT=\sqrt{a^{2012}+2011}+\dfrac{1}{\sqrt{a^{2012}+2011}}>=2\sqrt{\sqrt{a^{2012}+2011}\cdot\dfrac{1}{\sqrt{a^{2012}+2011}}}=2\)