如何解方程组(x+2y)(x+2z)=-16,(y+2z)(y+2x)=8,(z+2x)(z+2y)=-7
如何解方程组(x+2y)(x+2z)=-16,(y+2z)(y+2x)=8,(z+2x)(z+2y)=-7
方程组 2x+y+z=7,x+2y+z=8,x+y+2z=9 快
解方程组:2x+y+z=15 x+2y+z=16 x+y+z=17
解方程组2x+y-z=7 x+y+z =1 2x-y-z=5
x+z=y 7*x=y+z+2 x+y+z=14 解这个方程组、
已知x、y、z满足方程组:x+y-z=6;y+z-x=2;z+x-y=0 求x、y、z的值
解方程组:x:y:z=4:7:8,x+y+2z=54
解方程组 2x+y+z=7 x+2y+z=8 x+y+2x=9
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
解方程组{z=x+y 3x-2y-2z=-5 2x+y-z=3
解方程组:x-2y+z=-1,x+y+z=2,x+2y+3z=-1
解方程组 z=x+y 2x-3y+2z=5 x+2y-z=3