Do x+y+z=0x+y+z=0 nên ta có: x=−(y+z)y=−(x+z)z=−(x+z)x=−(y+z)y=−(x+z)z=−(x+z)
↔↔ x2=(y+z)2y2=(x+z)2z2=(x+z)2x2=(y+z)2y2=(x+z)2z2=(x+z)2
Thay vào ta được:
ax2+by2+cz2=a(y+z)2+b(x+z)2+c(x+y)2=a(y2+2yz+z2)+b(x2+2xz+z2)+c(x2+2xy+z2)ax2+by2+cz2=a(y+z)2+b(x+z)2+c(x+y)2=a(y2+2yz+z2)+b(x2+2xz+z2)+c(x2+2xy+z2)
=ay2+2ayz+az2+bx2+2bxz+bz2+cx2+2cxy+cy2=x2(b+c)+y2(a+c)+z2(a+b)+2(ayz+bxz+cxy)=ay2+2ayz+az2+bx2+2bxz+bz2+cx2+2cxy+cy2=x2(b+c)+y2(a+c)+z2(a+b)+2(ayz+bxz+cxy)
(*)
Lại có: a+b+c=0a+b+c=0 nên ta có: b+c=−aa+c=−ba+b=−c(1)b+c=−aa+c=−ba+b=−c(1)
ax+by+cz=0↔ayz+bxz+cxyxyz=0↔ayz+bxz+cxy=0(2)ax+by+cz=0↔ayz+bxz+cxyxyz=0↔ayz+bxz+cxy=0(2)
Thay (1),(2)(1),(2) và (*) ta được:
ax2+by2+cz2=x2(b+c)+y2(a+c)+z2(a+b)+2(ayz+bxz+cxy)=−ax2−by2−cz2+2.0=−ax2−by2−cz2ax2+by2+cz2=x2(b+c)+y2(a+c)+z2(a+b)+2(ayz+bxz+cxy)=−ax2−by2−cz2+2.0=−ax2−by2−cz2
↔ax2+by2+cz2=−ax2−by2−cz2↔ax2+by2+cz2=−ax2−by2−cz2
2 số đối nhau bằng nhau nên 2 số đó bằng 0
→ax2+by2+cz2=0(dpcm)
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
