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