Question

a+b+c=0 rut gon A=\(\dfrac{1}{b^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)

Answer 1

Vì a+b+c=0. Suy ra

* a+b=-c

=> (a+b)²=c²

=> a²+b²+2ab=c²

=>a²+b²-c²=-2ab

tương tự ta đc a²+c²-b²=-2ac và c²+b²-a²=-2bc

Ta có

A=\(\dfrac{1}{a^2+c^2-a^2}+\dfrac{1}{c^2+a^2-b^2}+\dfrac{1}{a^2+b^2-c^2}\)

=>\(A=\dfrac{-1}{2bc}-\dfrac{1}{2ac}-\dfrac{1}{2ab}\)

=>A=\(\dfrac{-a}{2abc}-\dfrac{b}{2abc}-\dfrac{c}{2abc}\)

=>A=\(\dfrac{-a-b-c}{2abc}=\dfrac{-\left(a+b+c\right)}{2abc}\)

=>\(\dfrac{0}{2abc}=0\) (vì a+b+c=0)

vậy A=0