$\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}(a,b,c,d>0)$

$\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=k$.

Suy ra

d = ka

c = kd = k$^{2}$a

b = kc = k$^{3}$a    (*)

a = kb               (**)

Từ (*) và (**) suy ra:

b =  k$^{3}$a= k$^{4}$b$\Rightarrow k=1$

Vậy a = b = c = d

Do đó:

$A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}+\frac{2011d-2010a}{b+c}$

$A=4.\frac{a}{2a}=2$