Điều kiện a,b,c khác 0
Vì a+b+c=0 nên a=-(b+c) =>a^2=(b+c)^2
b=-(a+c)=>b^2=(a+c)^2
c=-(a+b)=>c^2=(a+b)^2
Xét \(\dfrac{a^2}{a^2-b^2-c^2}\)ta có
\(\dfrac{a^2}{a^2-b^2-c^2}=\dfrac{a^2}{a^2+2bc-\left(b^2+c^2+2bc\right)}\)
\(=\dfrac{a^2}{a^2-\left(b+c\right)^2+2bc}=\dfrac{a^2}{2bc}\)\(=\dfrac{a^3}{2abc}\)(1)
Tương tự ta tính được
\(\dfrac{b^2}{b^2-a^2-c^2}=\dfrac{b^3}{2abc}\)(2)
\(\dfrac{c^2}{c^2-a^2-b^2}=\dfrac{c^3}{2abc}\)(3)
Từ (1);(2);(3)=>A=\(\dfrac{a^3+b^3+c^3}{2abc}\)
=\(\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\)
=\(\dfrac{a^3+b^3+3ab\left(a+b\right)-3abc-3ab\left(a+b\right)+3abc}{2abc}\)
=\(\dfrac{\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)+3abc}{2abc}\)
=\(\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)+3abc}{2ab}\)
=\(\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vìa+b+c=0)
Vậy A=\(\dfrac{3}{2}\)
