Bài 31:
a+b+c=0
=>b+c=-a; a+c=-b; a+b=-c
\(A=\dfrac{a^2}{a^2-b^2-c^2}+\dfrac{b^2}{b^2-c^2-a^2}+\dfrac{c^2}{c^2-a^2-b^2}\)
\(=\dfrac{a^2}{a^2-\left(b^2+c^2\right)}+\dfrac{b^2}{b^2-\left(c^2+a^2\right)}+\dfrac{c^2}{c^2-\left(a^2+b^2\right)}\)
\(=\dfrac{a^2}{a^2-\left[\left(b+c\right)^2-2bc\right]}+\dfrac{b^2}{b^2-\left[\left(c+a\right)^2-2ac\right]}+\dfrac{c^2}{c^2-\left[\left(a+b\right)^2-2ab\right]}\)
\(=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ac}+\dfrac{c^2}{c^2-c^2+2ab}\)
\(=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2ab}=\dfrac{a^3+b^3+c^3}{2abc}\)
\(=\dfrac{\left(a+b\right)^3-3ab\left(a+b\right)+c^3}{2abc}=\dfrac{\left(-c\right)^3+c^3-3ab\left(-c\right)}{2abc}\)
\(=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)