解x(x+y+z)=4-yz、y(x+y+z)=9-xz、z(x+y+z)=25-xy方程组

（一）x(x+y+z)=4-yz.===>x²+(y+z)x+yz=4.===>(x+y)(x+z)=4①.同理,将后面两个方程变形可得(x+y)(y+z)=9,②(x+z)(y+z)=25.③,将三个方程相乘可得(x+y)(y+z)(x+z)=±30.④.(二）当(x+y)(y+z)(x+z)=30时,与前面三个方程相除得y+z=15/2.x+z=10/3.x+y=6/5.这三个方程相加得x+y+z=361/60.再与前面三个方程先将得:x=-89/60.y=161/60.z=289/60.(三）当(x+y)(y+z)(z+x)=-30时,同理可解.