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C

cchhbibi

a+b+c=0~> a^2=(b+c)^2, b^2=(a+c)^2, c^2=(a+b)^2
Do đó a^2x+b^2y+c^2z=x(b+c)^2+y(a+c)^2+z(a+b)^2
=x(b^2+2bc+c^2)+y(a^2+2ac+c^2)+z(a^2+2ab+b^2)
=a^2y+a^2z+b^2x+b^2z+c^2x+c^2y+2bcx+2acy+2abz
=a^2(y+z)+b^2(x+z)+c^2(x+y)+2(bcx+acy+abz) (1)
Thay y+z=-x, x+z=-y, x+y=-z và bcx+acy+abz=0( do x/a+y/b+z/c=0 ) vào (1) ta có:
a^2x+b^2y+c^2z=-a^2x-b^2y-c^2z
~> 2(a^2x+b^2y+c^2z)=0 ~>a^2x+b^2y+c^2z=0
 
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