x+2y-3z=10 3x+2y=10 6y+2z=-1
三元一次方程组数学题x+2y+2z=33x+y-2z=72x+3y-2z=10x-y=2z-x=3y+z=-1x-y-z
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
x+y+z=4 2x+3y-z=6 3x+2y+2z=10
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
[3x+2y+z=14,x+y+z=10,2x+3y-z=1]
{2x+3y-4z=-5 x+y+z=6 x-y+3z=10
(x-2y+z)/9=(2x+y+3z)/10=-(3x+2y-4z)/3=1 连等,求x,y,z,
已知x^3+y^3-6x+2y+z+3+10=0,求(x-y+z)(x+y-z)的值.(z+3)是绝对值
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai
若4x-3y-6z=0,x+2y-7z=0,求代数式5x*5x+2y*2y-z*z/2x*2x-3y*3y-10z*10
2X+Y+Z=10,X+2Y+Z=-6,X+Y+2Z=8 5X+4Y-3Y=13,2X-3Y+7Z=19,3X-Y=2X
如果|x+y+z-6|+|2x+3y-z-12|+|2x-y-z|=0求x,y,