ta có x+y+z=0 =>x^2=(y+z)^2
y^2=(x+z)^2
z^2=(x+y)^2
do đó ax^2+by^2+cz^2
=a(y+z)^2+b(x+z)^2+c(x+y)^2
=a(y^2+2yz+z^2)+b(x^2+2xz+z^2)
+c(x^2+2xy+y^2)
=x^2(b+c)+y^2(a+c)+z^2(a+b)
+2(ayz+bxz+cxy) (1)
thay b+c=-a ,a+c=-b , a+b=-c do a+b+c=0
và ayz+bxz+cxy=0 do a/x+b/y+c/z=0 vào (1) ta được
ax^2+by^2+cz^2 = -(ax^2+by^2+cz^2)
=> ax^2+by^2+cz^2=0
Ta có:
\(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{a+b}{x+y}=0\Leftrightarrow ay\left(x+y\right)+bx\left(x+y\right)+xy\left(a+b\right)=0\)
\(\Leftrightarrow axy+ay^2+bx^2+bxy+axy+bxy=0\)
\(\Leftrightarrow ay^2+2axy+2bxy+bx^2=0\)
Vậy:
\(ax^2+by^2+cz^2=ax^2+by^2-\left(a+b\right)\left(x+y\right)^2\)
\(=ax^2+by^2-\left(ax^2+2axy+ay^2+bx^2+2bxy+by^2\right)\)
\(=-\left(ay^2+2axy+2bxy+by^2\right)=-0=0\)
