Question

解方程组xy+xz=8-x^2,yx+yz=12-y^2,zy+zx=-4-z^2

Answer 1

xy+xz=8-x² yx+yz=12-y² zy+zx=-4-z²

x(x+y+z)=8 y(x+y+z)=12 z(x+y+z)=-4

(x+y+z)²=8+12-4=16

x+y+z=±4

则①x+y+z=4时 x=4-y-z 代入yx+yz=12-y²和 zy+zx=-4-z²,则y=3 z=-1

则x=2,y=3,z=-1

②x+y+z=-4时同理得x=-2,y=-3,z=1