Áp dụng TC DTSBN ta có :

\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d\right)}=\frac{1}{2}\)

\(\Rightarrow\frac{a}{2b}=\frac{1}{2}\Rightarrow a=\frac{1}{2}.2b\Rightarrow a=b\) (1)

\(\Rightarrow\frac{b}{2c}=\frac{1}{2}\Rightarrow b=\frac{1}{2}.2c\Rightarrow b=c\) (2)

\(\Rightarrow\frac{c}{2a}=\frac{1}{2}\Rightarrow c=\frac{1}{2}.2a\Rightarrow c=a\) (3)

\(\Rightarrow\frac{d}{2a}=\frac{1}{2}\Rightarrow d=\frac{1}{2}.2a\Rightarrow d=a\) (4)

Từ (1);(2);(3):(4) \(\Rightarrow a=b=c=d\) .Thay vào A ta được :

\(A=\frac{2011a-2010a}{a+a}+\frac{2011a+2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)

\(=\frac{a}{2a}+\frac{4021a}{2a}+\frac{a}{2a}+\frac{a}{2a}=\frac{a+4021a+a+a}{2a}=\frac{4024a}{2a}=\frac{4024}{2}=2012\)

Vậy \(A=2012\)

Áp dụng dãy tỉ số bằng nhau ta có: \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2a+2b+2c+2d}=\frac{1}{2}\)

\(\Rightarrow a=b;b=c;c=d;d=a\) hay \(a=b=c=d\)

\(\Rightarrow A=\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)

 \(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2\left(a+b+c+d+\right)}=\frac{1}{2}\)=\(\frac{1}{2}\)

\(\Rightarrow2a=2b,2b=2c,2c=2d,2d=2a\) 

\(\Leftrightarrow a=b=c=d\)

\(\Rightarrow A=\frac{2011a-2010b}{c+d}+\frac{2011b-2010c}{a+d}+\frac{2011c-2010d}{a+b}=\frac{2011d-2010a}{b+c}\)

\(\Leftrightarrow A=\frac{2011a-2010a}{a+a}+\frac{2011b-2010b}{b+b}+\frac{2011c-2010c}{c+c}+\frac{2011d-2010d}{d+d}\)

\(\Leftrightarrow A=\frac{a}{2a}+\frac{b}{2b}+\frac{c}{2c}+\frac{d}{2d}\) 

\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)