x+y+z=0
=> x+y=-z
=> (x+y)2=z2
=>x2+2xy+y2=z2
=>2xy=z2-x2-y2
tương tự ta được
2yz=x2-y2-z2
2xz=y2-x2-z2
ta lại có
*x+y+z=0 => x+y=-z hay z=-(x+y)
* x3+y3+z3
=(x3+y3)-(x+y)3
=(x+y)(x2-xy+y2)-(x+y)3
=(x+y)[x2-xy+y2-(x+y)2]
=(x+y)(x2-xy+y2-x2-2xy-y2)
=(x+y)(-3xy)
=-z.(-3xy)
=3xyz
=> A=\(\dfrac{x^2}{2xy}+\dfrac{y^2}{2xz}+\dfrac{z^2}{2xy}=\dfrac{x^3+y^3+z^3}{2xyz}=\dfrac{3xyz}{2xyz}=\dfrac{3}{2}\)