﹛x+2y+2z=3 3x+y-2y=7 2x+3y-2z=10
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
若2x+5y+4z=6,3x+y-7z=-4,那么x+y-z的值是多少?
x/2=y/3=z/5 x+3y-z/x-3y+z
解方程组:2x-y+z=6.-3x+2y-z=-9.x+y-2z=-1求解
4x-3y-6z=0,x+2y-7z=0,xyz的积不等于0,求(x+y-z)/(x-y+2z)的值
x/2=y/3=z/4则2x+y-z/3x一2y+Z
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai
已知x,y,z 大于0,x+y+z=2,求证 xz/y(y+z)+zy/x(x+y)+yx/z(z+x)大于等于2/3
如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y,
已知,方程组:4x-3y-7z=0 x+2y=10z 则(x-y+z)÷(x+y+z)=——
解方程组2x-y+3z=3,3x+y-2z=﹣1,x+y+z=5,若先消去Z得到含Y,X的二元一次方程组是什么
2X+Y+Z=10,X+2Y+Z=-6,X+Y+2Z=8 5X+4Y-3Y=13,2X-3Y+7Z=19,3X-Y=2X