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

\(\Rightarrow\dfrac{a^{2012}+2012}{\sqrt{a^{2012}+2011}}>2\)

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

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

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

\(x= \dfrac{2011^3-1}{2011^2+2012} = \dfrac{(2011-1)(2011^2+2011+1)}{2011^2 + 2011 + 1} = 2010\)

\(y = \dfrac{2012^3+1}{2012^2-2011} = \dfrac{(2012+1)(2012^2-2012+1)}{2012^2-2012 + 1} = 2013\)

Suy ra:

 x + y = 2010 + 2013 = 4023

b)

\(B=\dfrac{2011^{2012}+1}{2011^{2012}-4}>\dfrac{2011^{2012}+1+3}{2011^{2012}-4+3}=\dfrac{2011^{2012}+4}{2011^{2012}-1}=A\)

Vậy B>A

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