Cho [tex]\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{b+a}=1[/tex] [tex]
Tính N=\frac{a^{2}}{b+c}+\frac{b^{2}}{a+c}+\frac{c^{2}}{a+b}[/tex]
Từ gt ta có:
$(a+b+c)(\dfrac a{b+c}+\dfrac b{a+c}+\dfrac c{a+b})=a+b+c
\\\Rightarrow \dfrac{a^2}{b+c}+\dfrac{a(b+c)}{b+c}+\dfrac{b^2}{a+c}+\dfrac{b(a+c)}{a+c}+\dfrac{c^2}{a+b}+\dfrac{c(a+b)}{a+b}=a+b+c
\\\Rightarrow \dfrac{a^2}{b+c}+a+\dfrac{b^2}{a+c}+b+\dfrac{c^2}{a+b}+c=a+b+c
\\\Rightarrow N=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}=0$