3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
3x+2y+z=14 ,x+y+z=10,z+2x+3y=1 5
x/2=y/3=z/5 x+3y-z/x-3y+z
试证明(x+y-2z)+(y+z-2x)+(z+x-2y)=3(x+y-2z)(y+z-2x)(z+x-2y)
[3x+2y+z=14,x+y+z=10,2x+3y-z=1]
{x+y+z=1;x+3y+7z=-1;z+5y+8z=-2
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
(4x-2y-z)-{5x[8y-2y-(x+y)]-x+(3y-10z)]=? kuai
3x+2y+z=14 x+y+2z+-3 2x+3y-z=1
已知2x+5y+4z=15.7x+y+3z=14 则x+y+z=?
{2x+3y-4z=-5 x+y+z=6 x-y+3z=10
x+y+z=2 x-3y+2z=1 2x+2y+z+5 要具体步骤..
3道高数题,1,函数F(x,y,z)=(e^x) * y * (z^2) ,其中z=z(x,y)是由x+y+z+xyz=