Với \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\). Khi đó:
+,
\(\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-a^2-c^2}+\dfrac{c^2}{c^2-a^2-b^2}\\ =\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(a^2+c^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\\ =\dfrac{a^2}{a^2-\left(b+c\right)^2+2bc}+\dfrac{b^2}{b^2-\left(a+c\right)^2+2ac}+\dfrac{c^2}{c^2-\left(a+b\right)^2+2ab}\\ =\dfrac{a^2}{a^2-\left(-a\right)^2+2bc}+\dfrac{b^2}{b^2-\left(-b\right)^2+2ac}+\dfrac{c^2}{c^2-\left(-c\right)^2+2ab}\\ =\dfrac{a^3}{2abc}+\dfrac{b^3}{2abc}+\dfrac{c^3}{2abc}\\ =\dfrac{\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}\\ =\dfrac{0-3.\left(-c\right).\left(-a\right).\left(-b\right)}{2abc}=\dfrac{3}{2}\)
+,
\(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\\ =\dfrac{1}{\left(b+c\right)^2-a^2-2bc}+\dfrac{1}{\left(c+a\right)^2-b^2-2ac}+\dfrac{1}{\left(a+b\right)^2-c^2-2ab}\\ =\dfrac{1}{\left(-a\right)^2-a^2-2bc}+\dfrac{1}{\left(-b\right)^2-b^2-2ac}+\dfrac{1}{\left(-c\right)^2-c^2-2ab}\\ =\dfrac{1}{-2bc}+\dfrac{1}{-2ac}+\dfrac{1}{-2ab}\\ =-\dfrac{a+b+c}{2abc}=0\)
#$\mathtt{Toru}$
