Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2(a+b+c+d)}=\frac{1}{2}\)
\(\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1\Leftrightarrow a=b=c=d\)
Do đó:
\(A=\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
Vậy \(A=2\)
Ta có: \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2\left(a+b+c+d\right)}=\dfrac{1}{2}\)
\(\Rightarrow a=b;b=c;c=d;d=a\)
\(A=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
\(A=\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}\)
\(A=\dfrac{c+c+c+c}{c+c}=2\)
Vậy ....................
