Ta có : \(\frac{a}{2b}\) = \(\frac{b}{2c}\) = \(\frac{c}{2d}\) =\(\frac{d}{2a}\) =\(\frac{a+b+c+d}{2\left(a+b+c+d\right)}\) =\(\frac{1}{2}\) ( a,b,c,d>0)
\(\Rightarrow\) \(\frac{a}{2b}\) =\(\frac{1}{2}\) \(\Rightarrow\) a=b (1) \(\frac{c}{2d}\) =\(\frac{1}{2}\)\(\Rightarrow\)c=d (3)
\(\frac{b}{2c}\) = \(\frac{1}{2}\) \(\Rightarrow\) b=c (2) \(\frac{d}{2a}\)=\(\frac{1}{2}\) \(\Rightarrow\) d=a(4)
Từ (1) ,(2) ,(3) và (4) \(\Rightarrow\)a=b=c=d (5)
Từ (5) ta thấy :a=b ,a=c ,a=d
\(\Rightarrow\)\(\frac{2011a-200b}{c+d}\) + \(\frac{2011b-2010c}{a+d}\) +\(\frac{2011c-2010d}{a+b}\) + \(\frac{2011d-2010a}{b+c}\)
= \(\frac{2011a-2010b}{a+a}\) + \(\frac{2011a-2010a}{a+a}\) + \(\frac{2011a-2010a}{a+a}\) + \(\frac{2011a-2010a}{a+a}\)
= \(\frac{2011a-2010a+2011a-2010a+2011a-2010a+2011a-2010a}{2a}\)
= \(\frac{a+a+a+a}{2a}\)= \(\frac{4a}{2a}\)=2
KL : \(\frac{a}{2b}\) = \(\frac{b}{2c}\) = \(\frac{c}{2d}\) = \(\frac{d}{2a}\)(a,b,c,d>0) thì A =2
