Từ a+b+c=2010
\(\Rightarrow\)a= 2010-(b+c)
\(\Rightarrow\)b= 2010-(c+a)
\(\Rightarrow\)c= 2010-(a+b)
Thay vào A, ta được:
A=\(\frac{2010-\left(b+c\right)}{b+c}\)+ \(\frac{2010-\left(c+a\right)}{c+a}\) + \(\frac{2010-\left(a+b\right)}{a+b}\)
A= \(\frac{2010}{b+c}\)+ \(\frac{2010}{c+a}\)+\(\frac{2010}{a+b}\)- 3
A= 2010( \(\frac{1}{b+c}\)+\(\frac{1}{c+a}\)+\(\frac{1}{a+b}\) ) -3
A= 2010. \(\frac{1}{10}\)-3
A=201-3
A= 198
Vậy A=198