Giải
[tex]A=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b} =a(\frac{a}{b+c}+1-1)+b(\frac{b}{c+a}+1-1)+c(\frac{c}{a+b}+1-1)[/tex]
[tex]=a(\frac{a+b+c}{b+c}-1)+b(\frac{a+b+c}{c+a}-1)+c(\frac{a+b+c}{a+b}-1)=\frac{a(a+b+c)}{b+c}+\frac{b(a+b+c)}{c+a}+\frac{c(a+b+c)}{a+b}-(a+b+c)[/tex]
[tex](a+b+c)(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})-(a+b+c)=(a+b+c)-(a+b+c)=0[/tex]
[tex]\Rightarrow A=\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}=0[/tex] (ĐPCM)