\(\dfrac{1}{a^2+b^2-c^2}+\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}\)
\(=\dfrac{1}{a^2+b^2-\left(-a-b\right)^2}+\dfrac{1}{b^2+c^2-\left(-b-c\right)^2}+\dfrac{1}{c^2+a^2-\left(-c-a\right)^2}\)
\(=\dfrac{1}{a^2+b^2-\left(a+b\right)^2}+\dfrac{1}{b^2+c^2-\left(b+c\right)^2}+\dfrac{1}{c^2+a^2-\left(c+a\right)^2}\)
\(=\dfrac{1}{a^2+b^2-a^2-2ab-b^2}+\dfrac{1}{b^2+c^2-b^2-2bc-c^2}+\dfrac{1}{c^2+a^2-c^2-2ac-a^2}\)
\(=\dfrac{1}{-2ab}+\dfrac{1}{-2bc}+\dfrac{1}{-2ac}\)
\(=\dfrac{c+a+b}{-2abc}=\dfrac{0}{-2abc}=0\)
ta có a+b+c=0=>a+b=-c =>(a+b)^2=c^2=> a^2+b^2=c^2-2ab =>a^2+b^2-c^2=-2ab
tương tự ta sẽ có
-1/2ab-1/2bc-1/2ac =-c/2abc- a/2abc- b/2abc =0 (vì a+b+c=0)
