ta có:
\(\frac{b^2-c^2}{\left(a+b\right).\left(a+c\right)}=\frac{b^2-a^2+a^2-c^2}{\left(a+b\right).\left(a+c\right)}=\frac{\left(b-a\right).\left(b+a\right)+\left(a-c\right).\left(a+c\right)}{\left(a+b\right).\left(a+c\right)}=\frac{b-a}{a+c}+\frac{a-c}{a+b}\left(1\right)\)
\(\frac{c^2-a^2}{\left(b+c\right).\left(b+a\right)}=\frac{c^2-b^2+b^2-a^2}{\left(b+c\right).\left(b+a\right)}=\frac{\left(c-b\right).\left(b+c\right)+\left(b-a\right).\left(a+b\right)}{\left(b+c\right).\left(b+a\right)}=\frac{c-b}{b+a}+\frac{b-a}{b+c}\left(2\right)\)
\(\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}=\frac{a^2-c^2+c^2-b^2}{\left(c+a\right).\left(c+b\right)}=\frac{\left(a-c\right).\left(a+c\right)+\left(c-b\right).\left(c+b\right)}{\left(c+a\right).\left(c+b\right)}=\frac{a-c}{c+b}+\frac{c-b}{c+a}\left(3\right)\)
từ (1),(2),(3)
\(\Rightarrow\frac{b^2-c^2}{\left(a+b\right).\left(a+c\right)}+\frac{c^2-a^2}{\left(b+c\right).\left(b+a\right)}+\frac{a^2-b^2}{\left(c+a\right).\left(c+b\right)}\)
\(=\frac{b-a}{a+c}+\frac{a-c}{a+b}+\frac{c-b}{a+b}+\frac{b-a}{b+c}+\frac{a-c}{c+b}+\frac{c-b}{c+a}=\frac{c-a}{a+c}+\frac{b-c}{b+c}+\frac{a-b}{a+b}\Rightarrowđpcm\)
